Question

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

Answer 1

Consider given the given intigration,

Let,

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

Put,

$t={\tan }^{-1}x\Rightarrow x=\tan t$

$dt=\frac{1}{1+x^2}dx$

$dx=(1+x^2)dt$

$I=\int \frac{\left(\tan t\right).t}{1+x^2}(1+x^2)dt$

$I=\int t.\tan tdt$ ……(1)

$\because \int t.\tan tdt=\int x\tan xdx$ …….(2) ( by integration properties)

Since equation (1),

$=t\int \tan tdt-\int 1.\log \sec tdt$

$=t.\log \sec t-[\log \sec t.t-\int \frac{1}{\sec t}.\sec t\tan t.tdt$

$I=t.\log \sec t-\log \sec t-\int t.\tan tdt$

$I=t.\log \sec x-\log \sec x-I$ (Since equation (2))

$2I=t.\log \sec t-\log \sec t$

now,

$2I={\tan }^{-1}x.\log \sec {\tan }^{-1}x-\log \sec {\tan }^{-1}x$

$I=\frac{{\tan }^{-1}x.\log \sec {\tan }^{-1}x-\log \sec {\tan }^{-1}x}{2}$

Hence, this is the answer.