Question

Solve the integral :$\int \frac{x}{e^{x^2+1}}dx$

Short answer

The required answer is$-\frac{2x}{e^x}-\frac{1}{e^x}(x^2+1)-2e^{-x}+C$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int \frac{x}{e^{x^2+1}}dx$.

Step 2. Solve the integration by parts .  

We have,

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

and

$$\begin{gathered} dv=e^{-x}dx \\ v=\int e^{-x}dx \\ v=-\frac{1}{e^x} \end{gathered}$$

The formula of integration by parts is $\int udv=uv-\int vdu$.

$$\begin{gathered} \int \frac{x}{e^{x^2+1}}dx \\ =\left[(x^2+1)\left(-\frac{1}{e^x}\right)-\int \left(-\frac{1}{e^x}\right)dx\right] \\ =\left(-\frac{1}{e^x}\left(x^2+1\right)-\int \left(-\frac{2x}{e^x}\right)dx\right) \\ =\left(-\frac{1}{e^x}\left(x^2+1\right)+2\int \left(\frac{x}{e^x}\right)dx\right) \end{gathered}$$

Step 3. Integration by parts. 

We have,

$$\begin{gathered} u=x \\ du=dx \end{gathered}$$

and

$$\begin{gathered} dv=e^{-x}dx \\ v=\int e^{-x}dx \\ v=-\frac{1}{e^x} \end{gathered}$$

So,

$$\begin{gathered} \left(-\frac{1}{e^x}(x^2+1)+2\left(x\left(-\frac{1}{e^x}\right)-\int \left(-\frac{1}{e^x}\right)dx\right)\right) \\ =\left(-\frac{1}{e^x}(x^2+1)+2\left(-\frac{x}{e^x}-\int \left(-\frac{1}{e^x}\right)dx\right)\right) \\ =-\frac{1}{e^x}(x^2+1)+2\int e^{-x}dx \\ =-\frac{2x}{e^x}-\frac{1}{e^x}\left(x^2+1\right)-2e^{-x}+c \end{gathered}$$