Question

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

Short answer

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

Step-by-step solution

Step 1. Given information. 

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

Step 2. Solve the integration by parts .  

We have,

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

and

role="math" localid="1648720139610" $$\begin{gathered} dv=e^{-x}dx \\ v=\int e^{-x}dx \\ v=-\frac{1}{e^x} \end{gathered}$$

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

role="math" localid="1648720506794" $$\begin{gathered} \int \frac{x^2+1}{e^x}dx \\ =\left[\left(x^2+1\right)\left(-\frac{1}{e^x}\right)-\int \left(-\frac{1}{e^x}\right)dx\right] \\ =\left[-\frac{1}{e^x}\left(x^2+1\right)-\int \left(-\frac{2x}{e^x}\right)dx\right] \\ =\left[-\frac{1}{e^x}\left(x^2+1\right)+2\int \left(\frac{x}{e^x}\right)dx\right] \end{gathered}$$

Step 3. Integration by parts.

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

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

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