Question
Evaluate the following integrals:
$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
✓
Answer
$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
$=\int_{0}^\limits{1}2\tan^{-1}\text{x dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-2\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-\Big[\log\big(1+\log\text{x}^2\big)\Big]$
$=2\frac{\pi}{4}-0-\log2+0$
$=\frac{\pi}{2}-\log2$
$=\int_{0}^\limits{1}2\tan^{-1}\text{x dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-2\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-\Big[\log\big(1+\log\text{x}^2\big)\Big]$
$=2\frac{\pi}{4}-0-\log2+0$
$=\frac{\pi}{2}-\log2$
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