Question
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$
$\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$
$=\int3\tan^{-1}(\text{x})\text{dx}$
$=3\int\big[\tan^{-1}(\text{x})\times1\big]\text{dx}$
$=3\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{1}{1+\text{x}^2}\times\text{x dx}\Big]$
$=3\text{x}\tan^{-1}\text{x}-3\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
Let
$1+\text{x}^2=\text{t}$$\Rightarrow2\text{x dx = dt}$
Then,
$\text{I}=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\int\frac{\text{dt}}{\text{t}}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|\text{t}|+\text{C}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|1+\text{x}^2|+\text{C}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$