Evaluate the following integrals: ^ -1 ( 3x-x^3 1-3x^2 )dx - Vidyadip

Question

Evaluate the following integrals:$\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$

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Answer

Let $\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,
$\text{I}=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\int\frac{\text{dt}}{\text{t}}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|\text{t}|+\text{C}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|1+\text{x}^2|+\text{C}$

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