Let I(x) = (x^2x+ x)/((x x+1)^2),dx.

Question

Let $I(x) = \displaystyle\int\frac{x\sec^2x+	an x}{(x	an x+1)^2}\,dx$. If $I(0)=0$, then $I\left(\dfrac\pi4ight)$ is equal to:

Accepted Answer

Write the integral as $I(x) = \displaystyle\int\frac{x^2\cdot\frac{\sec^2x}{x}+\tan x}{(x\tan x+1)^2}\,dx$. Splitting via $x\sec^2x+\tan x = \big[(x\tan x+1)-1\big]\cdot\text{(derivative structure)}+2x$, one obtains

$$I(x) = \frac{x^2-1}{x\tan x+1}+2\int\frac{2x}{x\tan x+1}\,dx$$

(more precisely, using $\int\frac{x\sec^2x+\tan x}{(x\tan x+1)^2}dx = \frac{-x}{x\tan x+1}+\int\frac{2}{x\tan x+1}dx$ type manipulation), which after using $I(0)=0$ to fix the constant, gives

$$I\left(\frac\pi4\right) = \log_e\frac{(\pi+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}$$

Solutions · 1