Question

The value of $\underset{1}{\overset{e}{\int }}\frac{x^3+1}{x^4+x^3+x^2+x}dx$ is

• A. $\frac{4+\pi }{4}-{\cot }^{-1}\left(\sqrt{1+e^2}\right)$
• B. $\frac{4+\pi }{4}-{\sec }^{-1}\left(\sqrt{1+e^2}\right)$
• C. $\frac{4-\pi }{4}-{\cos }^{-1}\left(\sqrt{1+e^2}\right)$
• D. $\frac{4-\pi }{4}-{\sec }^{-1}\left(\sqrt{1+e^2}\right)$

Answer 1

The correct option is B $\frac{4+\pi }{4}-{\sec }^{-1}\left(\sqrt{1+e^2}\right)$

$I=\underset{1}{\overset{e}{\int }}\frac{x^3+1}{x^4+x^3+x^2+x}dx$
$=\underset{1}{\overset{e}{\int }}\frac{(x+1)(x^2-x+1)}{x(x+1)(x^2+1)}dx$
$=\underset{1}{\overset{e}{\int }}\frac{(x^2+1)-x}{x(x^2+1)}dx$
$=\underset{1}{\overset{e}{\int }}[\frac{1}{x}-\frac{1}{x^2+1}]dx$
$=[\ln |x|-{\tan }^{-1}x{]}_1^e$
$=1-{\tan }^{-1}\left(e\right)+\frac{\pi }{4}$
$=\frac{\pi +4}{4}-{\sec }^{-1}\left(\sqrt{1+e^2}\right)$