Question

Integrate the following function: $\frac{(x+1)(x+log\ x)^2}{x}$

Answer 1

we have to integrate function , $\frac{(x+1)(x+log\ x)^2}{x}$

first of all, resolve the given function.

$\frac{(x+1)(x+logx)^2}{x}$

= $\left(\frac{x+1}{x}\right)(x+logx)^2$

= $\left(1+\frac{1}{x}\right)(x+logx)^2$

Let x + logx = P .....(1)

differentiating both sides,

dx + 1/x . dx = dP

=> (1 + 1/x) dx = dP ........(2)

now, $\int{\frac{(x+1)(x+logx)^2}{x}}\,dx=\int{\left(1+\frac{1}{x}\right)(x+logx)^2}\,dx$

from equations (1) and (2),

$\int{\left(1+\frac{1}{x}\right)(x+logx)^2}\,dx=\int{P^2}\,dP$

= $\left[\frac{P^3}{3}\right]+C$

now putting P = (x + logx)

= $\left[\frac{(x+logx)^3}{3}\right]+C$