Q. 41 - Calculus

Question

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

$\int x(2^{x^2}+1)\mathrm{d}x$

Step-by-Step Solution

Answer

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

Step 1. Given Information

Solving the given integrals. 

$\int x(2^{x^2}+1)\mathrm{d}x$

Step 2. Solving the given integral using substitution method.

Let

$$\begin{gathered} u=x^2+1 \\ \frac{du}{dx}=2x \\ du=2xdx \\ \frac{1}{2}du=xdx \end{gathered}$$

Step 3. This substitution changes the integral into

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