Finding Antiderivatives

In order to use Part 2 of the Fundamental Theorem of Calculus, an antiderivative of the integrand is needed, but sometimes it is not easy to find one. This section collects some of the information we already have about the general properties of antiderivatives and about antiderivatives of particular functions. It shows how to use the information you already have to find antiderivatives of more complicated functions and introduces a "change of variable" technique to make this job easier.

INDEFINITE INTEGRALS AND ANTIDERIVATIVES

Antiderivatives appear so often that there is a notation to indicate the antiderivative of a function:

Definition: $\int \mathrm{f}(x) \mathrm{dx}$, read as "the indefinite integral of $f$" or as "the antiderivatives of $f$," represents the collection (or family) of functions whose derivatives are $f$.

If $F$ is an antiderivative of $f$, then every member of the family $\int \mathrm{f}(x) \mathrm{d} \mathrm{x}$ has the form $\mathrm{F}(x)+\mathrm{C}$ for some constant $C$. We write $\int \mathrm{f}(x) \mathrm{dx}=\mathrm{F}(x)+\mathrm{C}$, where $C$ represents an arbitrary constant.

There are no small families in the world of antiderivatives: if $f$ has one antiderivative $F$, then $f$ has an infinite number of antiderivatives and every one of them has the form $\mathrm{F}(x)+\mathrm{C}$. This means that there are many ways to write a particular indefinite integral and some of them may look very different. You can check that $\mathrm{F}(x)=\sin ^{2}(x), \mathrm{G}(x)=-\cos ^{2}(x)$, and $\mathrm{H}(x)=2 \sin ^{2}(x)+\cos ^{2}(x)$ all have the same derivative $\mathrm{f}(x)=2 \sin (x) \cos (x)$, so the indefinite integral of $2 \sin (x) \cos (x), \int 2 \sin (x) \cos (x) \mathrm{dx}$, can be written in several ways: $\sin ^{2}(x)+\mathrm{C} \text {, or }-\cos ^{2}(x)+\mathrm{C}, \text { or } 2 \sin ^{2}(x)+\cos ^{2}(x)+\mathrm{C}$.

Practice 1:Verify that $\int 2 \tan (x) \cdot \sec ^{2}(x) \mathrm{d} x=\tan ^{2}(x)+\mathrm{C}$ and $\int 2 \tan (x) \cdot \sec ^{2}(x) \mathrm{d} x=\sec ^{2}(x)+\mathrm{C}$.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-5.7-Finding-Antiderivatives.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

The following properties are true of all antiderivative

General Properties of Antiderivatives

If f and g are integrable functions, then

1. Constant Multiple $\int \mathrm{k} \cdot \mathrm{f}(x) \mathrm{dx}=\mathrm{k} \cdot \int \mathrm{f}(x) \mathrm{dx}$

2. Sum $\int \mathrm{f}(x)+\mathrm{g}(x) \mathrm{dx}=\int \mathrm{f}(x) \mathrm{dx}+\int \mathrm{g}(x) \mathrm{dx}$

3. Difference $\int \mathrm{f}(x)-\mathrm{g}(x) \mathrm{dx}=\int \mathrm{f}(x) \mathrm{dx}-\int \mathrm{g}(x) \mathrm{dx}$

There are general rules for derivatives of products and quotients. Unfortunately, there are no easy general patterns for antiderivatives of products and quotients, and only one more general property will be added to the list (in Chapter 6).

We have already found antiderivatives for a number of important functions

Particular Antiderivative

1. Constant Function: $\int \mathrm{k} \mathrm{d} \mathrm{x}=\mathrm{k} x+\mathrm{C}$

2. Powers of x: $\begin{aligned} &\int x^{\mathrm{n}} \mathrm{dx}=\frac{x^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C} \quad \text { if } \mathrm{n} \neq-1 \\ &\int x^{-1} \mathrm{dx}=\int \frac{1}{x} \mathrm{~d} \mathrm{x}=\ln |x|+\mathrm{C} \end{aligned}$

Common special cases: $\begin{aligned} &\int \sqrt{x} \mathrm{dx}=\frac{2}{3} x^{3 / 2}+\mathrm{C} \\ &\int \frac{1}{\sqrt{x}} \mathrm{dx}=2 x^{1 / 2}+\mathrm{C}=2 \sqrt{x}+\mathrm{C} \end{aligned}$

3. Trigonometric Functions: $\begin{aligned} &\int \cos (x) \mathrm{dx}=\sin (x)+\mathrm{C} \quad \int \sin (x) \mathrm{d} \mathrm{x}=-\cos (x)+\mathrm{C} \\ &\int \sec ^{2}(x) \mathrm{d} \mathrm{x}=\tan (x)+\mathrm{C} \quad \int \csc ^{2}(x) \mathrm{d} x=-\cot (x)+\mathrm{C} \\ &\int \sec (x) \cdot \tan (x) \mathrm{dx}=\sec (x)+\mathrm{C} \quad \int \csc (x) \cdot \cot (x) \mathrm{dx}=-\csc (x)+\mathrm{C} \end{aligned}$

4. Exponential Functions: $\int \mathrm{e}^{x} \mathrm{dx}=\mathrm{e}^{x}+\mathrm{C}$

All of these antiderivatives can be verified by differentiating. The list of antiderivatives of particular functions will continue to grow in the coming chapters and will eventually include antiderivatives of additional trigonometric functions, the inverse trigonometric functions, logarithms, and others.

Antiderivatives are very sensitive to small changes in the integrand, and we need to be very careful. Fortunately, an antiderivative can always be checked by differentiating, so even though we may not find the correct antiderivative, we should be able to determine if an antiderivative is incorrect.

Example 1: We know one antiderivative of $\cos (x)$ is $\sin (x)$, so $\int \cos (x) \mathrm{dx}=\sin (x)+\mathrm{C}$. Find $(a) \int \cos (2 x+3) \mathrm{dx}, (b) \int \cos (5 x-7) \mathrm{dx}, and (c) \int \cos \left(x^{2}\right) \mathrm{dx}.$

Solution:

(a) Since $\sin (x)$ is an antiderivative of $\cos (x)$, it is reasonable to hope that $\sin (2 x+3)$ will be an antiderivative of $\cos (2 x+3)$. When we differentiate $\sin (2 x+3)$, however, we see that $\mathbf{D}(\sin (2 x+3))=\cos (2 x+3) \cdot 2$, exactly twice the result we want. Let's try again by modifying our "guess" and differentiating. Since the result of differentiating was twice the function we wanted, lets try one half the original function $\mathbf{D}\left(\frac{1}{2} \sin (2 x+3)\right)=\frac{1}{2} \cos (2 x+3) \cdot 2=\cos (2 x+3)$, exactly what we want. Therefore, $\int \cos (2 x+3) \mathrm{dx}=\frac{1}{2} \sin (2 x+3)+\mathrm{C}$.

(b) $\mathbf{D}(\sin (5 x-7))=\cos (5 x-7) \cdot \mathbf{5}$, but $\mathbf{D}\left(\frac{1}{5} \sin (5 x-7)\right)=\frac{1}{5} \cos (5 x-7) \cdot 5=\cos (5 x-7)$ so $\int \cos (5 x-7) \mathrm{dx}=\frac{1}{5} \sin (5 x-7)+\mathrm{C}$.

(c) $\mathbf{D}\left(\sin \left(x^{2}\right)\right)=\cos \left(x^{2}\right) \cdot 2 x$. It was easy enough in parts (a) and (b) to modify our "guesses" to eliminate the constants 2 and 5, but the $x$ is much harder to eliminate.

$\mathbf{D}\left(\frac{1}{2 x} \sin \left(x^{2}\right)\right)=\mathbf{D}\left(\frac{\sin \left(x^{2}\right)}{2 x}\right)$ requires the quotient rule (or the product rule), and

$\mathbf{D}\left(\frac{\sin \left(x^{2}\right)}{2 x}\right)=\frac{2 x \cdot \mathbf{D}\left(\sin \left(x^{2}\right)\right)-\sin \left(x^{2}\right) \cdot \mathbf{D}(2 x)}{(2 x)^{2}}=\cos \left(x^{2}\right)-\frac{\sin \left(x^{2}\right)}{2 x^{2}}$.

which is not the result we want. A fact which can be proved using more advanced mathematics is that $\cos \left(x^{2}\right)$ does not have an "elementary" antiderivative composed of polynomials, roots, trigonometric functions, or exponential functions or their inverses. The value of a definite integral of $\cos \left(x^{2}\right)$ could still be approximated as accurately as needed by using Riemann sums or one of the numerical techniques in Section 4.9 . The point of part (c) is that even a simple looking integrand can be very difficult. At this point, there is no easy way to tell.

One method for finding antiderivatives is to "guess" the form of the answer, differentiate your "guess" and then modify your original "guess" so its derivative is exactly what you want it to be. Parts (a) and (b) of the previous example illustrate that technique.

Example 2: We know $\int \sec ^{2}(x) \mathrm{d} \mathrm{x}=\tan (x)+\mathrm{C}$ and $\int \frac{1}{\sqrt{x}} \mathrm{~d} \mathrm{x}=2 \sqrt{x}+\mathrm{C}$. Find (a) $\int \sec ^{2}(3 x+7) d x$ and (b) $\int \frac{1}{\sqrt{5 x+3}} \mathrm{dx}$.

Solution:

(a) If we "guess" an answer of $\tan (3 x+7)$ and then differentiate it, we get $\left.\mathbf{D}(\tan (3 x+7))=\sec ^{2}(3 x+7) \cdot \mathbf{D}(3 x+7)\right)=\sec ^{2}(3 x+7) \cdot 3$ which is 3 times too large. If we divide our original guess by 3 and try again, we have

$\mathbf{D}\left(\frac{1}{3} \tan (3 x+7)\right)=\frac{1}{3} \mathbf{D}(\tan (3 x+7))=\frac{1}{3} \sec ^{2}(3 x+7) \cdot 3=\sec ^{2}(3 x+7)$ so $\int \sec ^{2}(3 x+7) \mathrm{dx}=\frac{1}{3} \tan (3 x+7)+\mathrm{C}$.

(b) If we "guess" an answer of $2 \sqrt{5 x+3}$ and then differentiate it, we get $\mathbf{D}(2 \sqrt{5 x+3})=2 \cdot \frac{1}{2}(5 x+3)^{-1 / 2} \mathbf{D}(5 x+3)=5 \cdot(5 x+3)^{-1 / 2}=\frac{5}{\sqrt{5 x+3}}$ which is 5 times too large. Dividing our guess by 5 and differentiating, we have

$\mathbf{D}\left(\frac{2}{5} \sqrt{5 x+3}\right)=\frac{1}{5} \cdot \mathbf{D}(2 \sqrt{5 x+3})=\frac{1}{5} \cdot \mathbf{5} \cdot(5 x+3)^{-1 / 2}=\frac{1}{\sqrt{5 x+3}}$ $\int \frac{1}{\sqrt{5 x+3}} \mathrm{~d} x=\frac{2}{5} \sqrt{5 x+3}+\mathrm{C}$.

Practice 2: Find $\int \sec ^{2}(7 x) \mathrm{dx} \text { and } \int \frac{1}{\sqrt{3 x+8}} \mathrm{dx}$.

The "guess and check" method is a very effective technique if you can make a good first guess, one that misses the desired result only by a constant multiple. In that situation just divide the first guess by the unwanted constant multiple. If the derivative of your guess misses by more than a constant multiple, then more drastic changes are needed. Sometimes the next technique can help.

Successful integration is mostly a matter of recognizing patterns. The "change of variable" technique can make some underlying patterns of an integral easier to recognize. Basically, the technique involves rewriting an integral which is originally in terms of one variable, say $x$, in terms of another variable, say $u$. The hope is that after doing the rewriting, it will be easier to find an antiderivative of the new integrand.

For example, $\int \cos (5 x+1) \mathrm{dx}$ can be rewritten by setting $u=5 x+1$. Then $\cos (5 x+1)$ becomes $\cos (u)$ and $\mathbf{d u}=\mathrm{d}(5 x+1)=\mathbf{5} \mathrm{d} \mathrm{x}$ so the differential $dx$ is $\frac{1}{5} \mathrm{du}$. Finally $\int \cos (5 x+1) \mathrm{dx}=\int \cos (u) \cdot \frac{\mathbf{1}}{\mathbf{5}} \mathrm{du} =\frac{1}{5} \sin (u)+C=\frac{1}{5} \sin (5 x+1)+\mathrm{C}$.

The steps of the "change of variable" method can be summarized as

(1) set a new variable, say $u$, equal to some function of the original variable $x$ (usually $u$ is set equal to some part of the original integrand function)

(2) calculate the differential $du$ as a function of $dx$

(3) rewrite the original integral in terms of $u$ and $du$

(4) integrate the new integral to get an answer in terms of $u$

(5) replace the u in the answer to get an answer in terms of the original variable

Changing the Variable

Fig. 1 shows the general flow of this algorithm and another example.

Example 3: Make the suggested change of variable and rewrite each integral in terms of $u$ and $du$.

(a) $\int \cos (x) \mathrm{e}^{\sin (x)} \mathrm{dx} \text { with } u=\sin (x)(\mathrm{b})$

(b) $\int \frac{2 x}{5+x^{2}} \mathrm{dx} \text { with } u=5+x^{2}$

Solution:

(a) Since $u=\sin (x)$, then $\mathrm{du}=\mathrm{d}(\sin (x))=\cos (x) \mathrm{d} \mathrm{x}$ so $\mathrm{e}^{\sin (x)}=\mathrm{e}^{u}$ and $\cos (x) d x=d u$. Then $\int \cos (x) \mathrm{e}^{\sin (x)} \mathrm{dx}=\int \mathrm{e}^{u} \mathrm{du}=\mathrm{e}^{u}+\mathrm{C}=\mathrm{e}^{\sin (x)}+\mathrm{C}$.

(b) If $u=5+x^{2}$, then $\mathbf{d u}=\mathrm{d}\left(5+x^{2}\right)=\mathbf{2} x \mathbf{d} \mathbf{x}$, so $\int \frac{2 x}{5+x^{2}} \mathbf{d x}=\int \frac{1}{u} \mathbf{d u}=\ln |u|+C=\ln 5+x^{2} \mid+C$

In each example, the change of variable did not find the antiderivative, but it did make the pattern of the integrand more obvious so it was easier to determine an antiderivative.

Practice 3: Make the suggested change of variable and rewrite each integral in terms of $u$ and $du$.

(a) $\int(7 x+5)^{3} \mathrm{dx}$ with $u=7 x+5$

(b) $\int \sin \left(x^{3}-1\right) 3 x^{2} \mathrm{dx}$ with $u=x^{3}-1$

In the future, you will need to decide what u should equal, and there are no rules which guarantee that your choice will lead to an easier integral. There is, however, a "rule of thumb" which frequently results in easier integrals. Even though the following suggestion is not guaranteed, it is often worth trying.

A "Rule of thumb" for changing the variable

If part of the integrand is a composition of functions, $\mathrm{f}(\mathrm{g}(x))$,

then try setting $u=g(x)$, the "inner" function.

If part of the integrand is being raised to a power, try setting u equal to the part being raised to the power. For example, if the integrand includes $(3+\sin (x))^{5}$, try $u=3+\sin (x)$. If part of the integrand involves a trigonometric (or exponential or logarithmic) function of another function, try setting $u$ equal to the "inside" function. If the integrand includes the function $\sin \left(3+x^{2}\right), \text { try } u=3+x^{2}$.

The key to becoming skilled at selecting a good u and correctly making the substitution is practice.

Example 4: Select a function for u for each integral and rewrite the integral in terms of $u$ and $du$.

Solution:

(a) Put $u=2+\sin (3 x)$. Then $\mathrm{du}=3 \cos (3 x) \mathrm{dx}$, and the integral becomes $\int \frac{1}{3} \sqrt{u} \mathrm{du}$.

(b) Put $u=2+\mathrm{e}^{x}$. Then $\mathrm{du}=\mathrm{e}^{\boldsymbol{x}} \mathrm{d} \mathbf{x}$, and the integral becomes $\int \frac{5}{u} \mathrm{du}$.

(c) Put $u=\mathrm{e}^{x}$. Then $\mathrm{du}=\mathrm{e}^{\boldsymbol{x}} \mathrm{dx}$, and the integral becomes $\int \sin (u) \mathrm{du}$

Once an antiderivative in terms of $u$ is found, we have a choice of methods. We can

(a) rewrite our antiderivative in terms of the original variable $x$ , and then evaluate the antiderivative at the integration endpoints and subtract, or

(b) change the integration endpoints to values of $u$, and evaluate the antiderivative in terms of $u$ before subtracting.

If the original integral had endpoints $x=\mathrm{a}$ and $x=\mathrm{b}$, and we make the substitution $u=\mathrm{g}(x)$ and $\mathrm{du}=g^{\prime}(x) \mathrm{d} \mathrm{x}$, then the new integral will have endpoints $u=\mathrm{g}(\mathrm{a})$ and $u=\mathrm{g}(\mathrm{b})$ and

$\int_{x=\mathrm{a}}^{x=\mathrm{b}}$ (original integrand) $dx$ becomes $\int_{u=g(a)}^{u=g(b)}$ (new integrand) $du$.

To evaluate $\int_{0}^{1}(3 x-1)^{4} \mathrm{~d} \mathrm{x}$, we can put $u=3 x-1$. Then $\mathbf{d u}=\mathrm{d}(3 x-1)=\mathbf{3} \mathbf{d} \mathbf{x}$ so the integral becomes $\int u^{4} \frac{\mathbf{1}}{\mathbf{3}} \mathrm{du}=\frac{1}{15} u^{5}+\mathrm{C}$.

(a) Converting our antiderivative back to the variable $x$ and evaluating with the original endpoints:

$\int_{0}^{1}(3 x-1)^{4} \mathrm{dx}=\left.\frac{1}{15}(3 x-1)^{5}\right|_{0} ^{1}=\frac{1}{15}(3 \cdot 1-1)^{5}-\frac{1}{15}(3 \cdot 0-1)^{5}=\frac{1}{15}(2)^{5}-\frac{1}{15}(-1)^{5}=\frac{32}{15}-\frac{-1}{15}=\frac{33}{15}$.

(b) Converting the integration endpoints to values of $u$: when $\mathrm{x}=0$, then $u=3 x-1=3 \cdot 0-1=-1$, and when $\mathrm{x}=1$, then $u=3 x-1=3 \cdot 1-1=2$ so

$\int_{x=0}^{x=1}(3 x-1)^{4} \mathrm{dx}=\int_{u=-1}^{u=2} u^{4} \frac{1}{3} \mathrm{du}=\left.\frac{1}{15} \mathrm{u}^{5}\right|_{-1} ^{2}=\frac{1}{15}(2)^{5}-\frac{1}{15}(-1)^{5}=\frac{32}{15}-\frac{-1}{15}=\frac{33}{15}$.

Both approaches typically involve about the same amount of work and calculation.

Practice 4: If the original integrals in Example 4 had endpoints (a) $\mathrm{x}=0$ to $\mathrm{x}=\pi$, (b) $\mathrm{x}=0$ to $\mathrm{x}=2$, and (c) and $\mathrm{x}=0$ to $x=\ln (3)$, then the new integrals should have what endpoints?

Special Transformations for $\int \sin ^{2}(x) \mathrm{dx} \text { and } \int \cos ^{2}(x) \mathrm{dx}$

The integrals of $\sin ^{2}(x)$ and $\cos ^{2}(x)$ occur relatively often, and we can find their antiderivatives with the help of two trigonometric identities for $\cos (2 x)$:

$\cos (2 x)=1-2 \sin ^{2}(x) \text { and } \cos (2 x)=2 \cos ^{2}(x)-1$

Solving the identies for $\sin ^{2}(x)$ and $\cos ^{2}(x)$, we get $\sin ^{2}(x)=\frac{1-\cos (2 x)}{2}$ and $\cos ^{2}(x)=\frac{1+\cos (2 x)}{2}$

Then $\begin{aligned} \int \sin ^{2}(x) \mathrm{dx} &=\int \frac{1-\cos (2 x)}{2} \mathrm{dx}=\int \frac{1}{2} \mathrm{~d} x-\int \frac{1}{2} \cos (2 x) \mathrm{d} x \\ &=\frac{x}{2}-\frac{1}{2} \frac{\sin (2 x)}{2}+\mathrm{C}=\frac{x}{2}-\frac{\sin (2 x)}{4}+\mathrm{C} \text { or } \frac{1}{2}\{x-\sin (x) \cos (x)\}+\mathrm{C} \end{aligned}$

Similarly, $\begin{aligned} \int \cos ^{2}(x) \mathrm{dx} &=\int \frac{1+\cos (2 x)}{2} \mathrm{dx}=\int \frac{1}{2} \mathrm{~d} x+\int \frac{1}{2} \cos (2 x) \mathrm{d} x \\ &=\frac{x}{2}+\frac{1}{2} \frac{\sin (2 x)}{2}+\mathrm{C}=\frac{x}{2}+\frac{\sin (2 x)}{4}+\mathrm{C} \text { or } \frac{1}{2}\{x+\sin (x) \cos (x)\}+\mathrm{C} \end{aligned}$

Practice 1: $\begin{aligned} &\mathbf{D}\left(\tan ^{2}(x)+\mathrm{C}\right)=2 \tan ^{1}(x) \mathbf{D}(\tan (x))=2 \tan (x) \sec ^{2}(x) \\ &\mathbf{D}\left(\sec ^{2}(x)+\mathrm{C}\right)=2 \sec ^{1}(x) \mathbf{D}(\sec (x))=2 \sec (x) \sec (x) \tan (x)=2 \tan (x) \sec ^{2}(x) \end{aligned}$

Practice 2: We know $\mathbf{D}(\tan (x))=\sec ^{2}(x)$, so it is reasonable to try $\tan (7 x)$.

$\mathbf{D}(\tan (7 x))=\sec ^{2}(7 x) \mathbf{D}(7 x)=7 \cdot \sec ^{2}(7 x)$, a result 7 times the function we want, so

divide the original "guess" by 7 and try it. $\mathrm{D}\left(\frac{1}{7} \tan (7 x)\right)=\frac{1}{7} \cdot 7 \cdot \sec ^{2}(7 x)=\sec ^{2}(7 x)$.

$\int \sec ^{2}(7 x) \mathrm{dx}=\frac{1}{7} \tan (7 x)+\mathrm{C}$

$\mathbf{D}\left((3 x+8)^{1 / 2}\right)=\frac{1}{2} \cdot(3 x+8)^{-1 / 2} \mathbf{D}(3 x+8)=\frac{3}{2} \cdot(3 x+8)^{-1 / 2}$ so let's multiply our original "guess" by 2/3:

$\mathbf{D}\left(\frac{2}{3}(3 x+8)^{1 / 2}\right)=\frac{2}{3} \frac{1}{2} \cdot(3 x+8)^{-1 / 2} \mathbf{D}(3 x+8)=\frac{2}{3} \frac{3}{2} \cdot(3 x+8)^{-1 / 2}=(3 x+8)^{-1 / 2}$

$\int \frac{1}{\sqrt{3 x+8}} \mathrm{dx}=\frac{2}{3}(3 x+8)^{1 / 2}+\mathrm{C}$.

Practice 3:

(a) $\int(7 x+5)^{3} \mathrm{dx}$ with $u=7 x+5$. Then $\mathrm{du}=7 \mathrm{dx}$ so $\mathrm{dx}=\frac{1}{7} \mathrm{du}$ and $\int(7 x+5)^{3} \mathrm{dx}=\int \mathrm{u}^{3} \frac{1}{7} \mathrm{du}=\frac{1}{7} \cdot \frac{1}{4} \mathrm{u}^{4}+\mathrm{C}=\frac{1}{28}(7 x+5)^{4}+\mathrm{C}$.

(b)$\int \sin \left(x^{3}-1\right) 3 x^{2} \mathrm{dx}$ with $u=x^{3}-1$. Then $\mathrm{du}=3 x^{2} \mathrm{dx}$ so

$\int \sin \left(x^{3}-1\right) 3 x^{2} \mathrm{dx}=\int \sin (u) \mathrm{du}=-\cos (u)+\mathrm{C}=-\cos \left(x^{3}-1\right)+\mathrm{C}$.

Practice 4:

(a) $\mathrm{u}=2+\sin (3 \mathrm{x})$ so when $\mathrm{x}=0, \mathrm{u}=2+\sin (3+0)=2$. When $\mathrm{x}=\pi$, $\mathrm{u}=2+\sin (3 \cdot \pi)=2$. (This integral is now very easy to evaluate. Why?)

(b) $\mathrm{u}=2+\mathrm{e}^{\mathrm{x}}$ so when $\mathrm{x}=0, \mathrm{u}=2+\mathrm{e}^{0}=3$. When $\mathrm{x}=2, \mathrm{u}=2+\mathrm{e}^{2}$.

The new integration endpoints are $\mathrm{u}=3$ to $\mathrm{u}=2+\mathrm{e}^{2}$.

(c) $\mathrm{u}=\mathrm{e}^{\mathrm{x}}$ so when $\mathrm{x}=0, \mathrm{u}=\mathrm{e}^{0}=1$. When $\mathrm{x}=\ln (3), \mathrm{u}=\mathrm{e}^{\ln (3)}=3$.

The new integration endpoints are $\mathbf{u}=1$ to $\mathbf{u}=3$.