Evaluate the following integrals: x^2 ^ -1 x 1+x^2 dx

Question

Evaluate the following integrals:

$\int\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\text{dx}$

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Answer

Let $\text{I}=\int\Big(\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\int\Big(\frac{\text{x}^2+1-1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int\Big(1-\frac{1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int1.\tan^{-1}\text{x dx}-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\int1\text{dx}\Big\}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
Putting $\text{x}^2+1=\text{t}$ in the first integral and $\tan^{-1}\text{x}=\text{p}$ in the second integral
$\Rightarrow2\text{x dx = dt} $ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\therefore \text{I}=\tan^{-1}\text{x.x}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}}-\int\text{p.dp}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|\text{t}|-\frac{\text{P}^2}{2}+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|1+\text{x}^2|-\frac{(\tan^{-1}\text{x})^2}{2}+\text{C}$ $[\therefore \text{t}=\text{x}^2+1\text{ and}\text{ p}=\tan^{-1}\text{x}]$

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