Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int x\ln \left(e^{x^2+1}\right)\mathrm{d}x$

Short answer

The solution of the given integral is $\int x\ln \left(e^{x^2+1}\right)\mathrm{d}x=\frac{1}{2}{e}^{\ln \left(e^{x^2+1}\right)}+C$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int x\ln \left(e^{x^2+1}\right)\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

$$\begin{gathered} u=\ln \left(e^{x^2+1}\right) \\ \frac{du}{dx}=\frac{2x}{e^{x^2+1}} \\ du=\frac{2x}{e^{x^2+1}}dx \\ \frac{e^{x^2+1}}{2}du=xdx \end{gathered}$$

Step 3. In this substitution after differentiation not at all in the form of f'(u(x))u'(x).

A clever change of variables will allow us to rewrite the integral so that it can be algebraically simplified.

$$\begin{gathered} u=\ln \left(e^{x^2+1}\right) \\ {e}^u=e^{x^2+1} \end{gathered}$$

$\frac{e^u}{2}du=xdx$

Step 4. This substitution changes the integral into 

$$\begin{gathered} \int x\ln \left(e^{x^2+1}\right)\mathrm{d}x=\frac{1}{2}\int e^{u}du \\ \int x\ln \left(e^{x^2+1}\right)\mathrm{d}x=\frac{1}{2}{e}^u+C \\ \int x\ln \left(e^{x^2+1}\right)\mathrm{d}x=\frac{1}{2}{e}^{\ln \left(e^{x^2+1}\right)}+C \end{gathered}$$