Question

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

• A. ${-e}^{{tan}^{-1}x}+c$
• B. $e^{{tan}^{-1}x}+c$
• C. ${-xe}^{{tan}^{-1}x}+c$
• D. ${xe}^{{tan}^{-1}x}+c$

Answer 1

Answer: (D)

${xe}^{{tan}^{-1}x}+c$

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

Let $x=\tan t\to dx={\sec }^{2}tdt$

$=\int e^{{\tan }^{-1}\left(\tan t\right)}\left(\frac{1+\tan t+{\tan }^{2}t}{1+{\tan }^{2}t}\right){\sec }^{2}tdt$

$=\int e^t\left(\frac{\tan t+{\sec }^{2}t}{{\sec }^{2}t}\right){\sec }^{2}tdt$

$=\int e^t(\tan t+{\sec }^{2}t)dt$

$=\int e^t\tan tdt+\int e^t{\sec }^{2}tdt$

$\tan t\int e^{t}dt-\int \frac{d}{dt}\left(\tan t\right)\left(\int e^{t}dt\right)dt+\int e^t\left({\sec }^{2}t\right)dt+c$

$=e^t\tan t-\int e^t\left({\sec }^{2}t\right)dt+\int e^t\left({\sec }^{2}t\right)dt+c$

$=e^t\tan t+c$

$=e^{{\tan }^{-1}x}\tan {\tan }^{-1}x+c$

$=x{e}^{{\tan }^{-1}x}+c$

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