Question

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

$\int 2x{e}^{3x^2}\mathrm{d}x$

Short answer

The solution of the given integral is $\int 2x{e}^{3x^2}\mathrm{d}x==\frac{1}{3}{\mathrm{e}}^{3{\mathrm{x}}^2}+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int 2x{e}^{3x^2}\mathrm{d}x$

Step 2. Solving the given integral using substitution method.  

Let

$$\begin{gathered} u=3x^2 \\ \frac{du}{dx}=6x \\ du=6xdx \\ \frac{1}{6}du=xdx \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int 2x{e}^{3x^2}\mathrm{d}x=\frac{2}{6}\int e^u\mathrm{du} \\ \int 2{\mathrm{xe}}^{3{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{3}{\mathrm{e}}^{\mathrm{u}}+\mathrm{C} \\ \int 2{\mathrm{xe}}^{3{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{3}{\mathrm{e}}^{3{\mathrm{x}}^2}+\mathrm{C} \end{gathered}$$