Question
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
✓
Answer
Let $\text{I}=-\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)\sec^2\theta\text{d}\theta$
$=\int\tan^{-1}(\tan2\theta)\sec^2\theta\text{d}\theta$
$=\int2\theta\sec^2\theta\text{d}\theta$
$=2\Big[\theta\int\sec^2\theta\text{d}\theta-\int(1\int\sec^2\theta\text{d}\theta)\text{d}\theta\Big]$
$=2\big[\theta\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta-\log\sec\theta\big]+\text{c}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log\big|1+\text{x}^2\big|+\text{C}$
Let $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)\sec^2\theta\text{d}\theta$
$=\int\tan^{-1}(\tan2\theta)\sec^2\theta\text{d}\theta$
$=\int2\theta\sec^2\theta\text{d}\theta$
$=2\Big[\theta\int\sec^2\theta\text{d}\theta-\int(1\int\sec^2\theta\text{d}\theta)\text{d}\theta\Big]$
$=2\big[\theta\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta-\log\sec\theta\big]+\text{c}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log\big|1+\text{x}^2\big|+\text{C}$
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