Question

Integrate the function.
$\int xta{n}^{-1}xdx.$

Answer 1

Let $I=\int xta{n}^{-1}xdx$
On taking $ta{n}^{-1}x$ as first function and x as second function and integrating by parts, we get
$(\because$ Inverse functions comes before algebraic function in ILATE)
$\therefore I=ta{n}^{-1}x\int xdx-\int \left[\frac{d}{dx}\right(ta{n}^{-1}x)\int xdx]dx$
$=ta{n}^{-1}x.\frac{x^2}{2}-\int [\frac{1}{1+x^2}.\frac{x^2}{2}]dx=\frac{x^2}{2}.ta{n}^{-1}x-\frac{1}{2}\int \left[\frac{x^2+1-1}{1+x^2}\right]dx$
[Add and subtract 1 in numerator of second term]
$=\frac{x^2}{2}.ta{n}^{-1}x-\frac{1}{2}[\int \frac{1+x^2}{1+x^2}dx-\int \frac{1}{1+x^2}dx]=\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}[\int 1dx-\int \frac{1}{1+x^2}dx]=\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}x+\frac{1}{2}ta{n}^{-1}x+C=\left(\frac{x^2+1}{2}\right)ta{n}^{-1}x-\frac{1}{2}x+C$