Let f(x)=( x)/((1+x)^2),dx (x0).

Question

Let $f(x)=\displaystyle\int\frac{\sqrt x}{(1+x)^2}\,dx$ $(x\ge0)$. Then $f(3)-f(1)$ is equal to :

Accepted Answer

Put $x=\tan^2\theta\Rightarrow dx=2\tan\theta\sec^2\theta\,d\theta$:
$$f(x)=\int\frac{2\tan^2\theta\sec^2\theta}{\sec^4\theta}d\theta=2\int\sin^2\theta\,d\theta=\theta-\frac{\sin2\theta}2+C$$

With $\theta=\tan^{-1}\sqrt x$ and $\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}=\dfrac{2\sqrt x}{1+x}$:
$$f(x)=\tan^{-1}\sqrt x-\frac{\sqrt x}{1+x}+C$$

$$f(3)-f(1)=\tan^{-1}\sqrt3-\frac{\sqrt3}{4}-\tan^{-1}1+\frac12=\frac{\pi}3-\frac{\sqrt3}4-\frac\pi4+\frac12=\frac{\pi}{12}+\frac12-\frac{\sqrt3}4$$

Solutions · 1