Question

The integral $\int e^{ta{n}^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)$

• A. $\frac{e^{ta{n}^{-1}x}}{1+x^2}+C$
• B. $x{e}^{ta{n}^{-1}x}+C$
• C. $\frac{x{e}^{ta{n}^{-1}x}}{1+x^2}+C$
• D. none of these

Answer 1

The correct option is B $x{e}^{ta{n}^{-1}x}+C$

$\int e^{{\tan }^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)dx$

putting

$\begin{array}{c}{\tan }^{-1}x=t\\ x=\tan t\\ \frac{1}{1+x^2}dx=dt\\ \int e^t\left(1+\tan t+{\tan }^{2}t\right)dt\\ \int e^t\left(1+{\tan }^{2}t+\tan t\right)dt\\ \int e^t\left(\tan t+{\sec }^{2}t\right)dt\\ e^t\tan t+c\\ e^{{\tan }^{-1}x}\times x+c\\ x{e}^{{\tan }^{-1}x}+c\end{array}$

$B$ is the correct answer