Question

Review of integration by substitution: Use u-substitution to find each of the following integrals.

a. $\int e^{3x+1}dx$b. $\int x{e}^{x^2+1}dx$

c. $\int \frac{\ln x}{x}dx$d. $\int \frac{1}{x\ln x}dx$

e. $\int \tan x{\sec }^{2}xdx$f. $\int \frac{1}{x^2+1}dx$

g.$\int \sin x\sqrt{\cos x}dx$h.$\int e^x\sin e^{x}dx$

Short answer

Part (a)$\int e^{3x+1}dx=\frac{e^{3x+1}}{3}+C$

Part (b)$\int x{e}^{x^2+1}dx=\frac{e^{x^2+1}}{2}+C$

Part (c)$\int \frac{\ln x}{x}dx=\frac{{\left(\ln x\right)}^2}{2}+C$

Part (d)$\int \frac{1}{x\ln x}dx=\ln \left|\ln x\right|+C$

Part (e)$\int \tan x{\sec }^{2}xdx=\frac{{\tan }^{2}x}{2}+C$

Part (f)$\int \frac{1}{x^2+1}dx={\tan }^{-1}x+C$

Part (g)$\int \sin x\sqrt{\cos x}dx=-\frac{2{\cos }^{{\displaystyle \frac{3}{2}}}x}{3}+C$

Part (h)$\int e^x\sin e^{x}dx=-\cos \left(e^x\right)+C$

Step-by-step solution

Part (a) Step 1. Calculating the integral 

Given integral, $\int e^{3x+1}dx$

Substituting, $3x+1=u,3dx=du$

$$\begin{gathered} \int \frac{e^u}{3}du=\frac{e^u}{3}+C \\ =\frac{e^{3x+1}}{3}+C \end{gathered}$$

Part (b) Step 1. Calculating the integral 

Given integral, $\int x{e}^{x^2+1}dx$

Substituting, $x^2+1=u,2xdx=du$

$$\begin{gathered} \int \frac{e^u}{2}du=\frac{e^u}{2}+C \\ =\frac{e^{x^2+1}}{2}+C \end{gathered}$$

Part (c) Step 1. Calculating the integral 

Given integral, $\int \frac{\ln x}{x}dx$

Substituting, $\ln x=u,\frac{dx}{x}=du$

$$\begin{gathered} \int udu=\frac{u^2}{2}+C \\ =\frac{{\left(\ln x\right)}^2}{2}+C \end{gathered}$$

Part (d) Step 1. Calculating the integral 

Given integral, $\int \frac{1}{x\ln x}dx$

Substituting, $\ln x=u,\frac{dx}{x}=du$

$$\begin{gathered} \int \frac{1}{u}du=\ln \left|u\right|+C \\ =\ln \left|\ln x\right|+C \end{gathered}$$

Part (e) Step 1. Calculating the integral 

Given integral, $\int \tan x{\sec }^{2}xdx$

Substituting, $\tan x=u,se{c}^{2}xdx=du$

$$\begin{gathered} \int udu=\frac{u^2}{2}+C \\ =\frac{{\tan }^{2}x}{2}+C \end{gathered}$$

Part (f) Step 1. Calculating the integral 

Given integral, $\int \frac{1}{x^2+1}dx$

$={\tan }^{-1}x+C$

Part (g) Step 1. Calculating the integral 

Given integral, $\int \sin x\sqrt{\cos x}dx$

Substituting, $\cos x=u,-\sin xdx=du$

$$\begin{gathered} -\int \sqrt{u}du=-\frac{u^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}+C \\ =-\frac{2{\cos }^{{\displaystyle \frac{3}{2}}}x}{3}+C \end{gathered}$$

Part (h) Step 1. Calculating the integral 

Given integral, $\int e^x\sin e^{x}dx$

Substituting, $e^x=u,e^{x}dx=du$

$$\begin{gathered} \int \sin udu=-\cos u+C \\ =-\cos \left(e^x\right)+C \end{gathered}$$