Question

What is `int (lnx)^2 / x^3dx`?

Answer 1

`I=-1/(4x^2)[2(lnx)^2+2lnx+1]+c`

Explanation:

Here,

`I=int(lnx)^2/x^3 dx=int(lnx)^2/x^2*1/xdx`

Let,

`lnx=u=>1/xdx=du`

`and x=e^u`

So,

`I=intu^2/(e^u)^2*du`

`=int u^2*e^(-2u)du`

`"Using "color(blue)"Integration by Parts :"`

`I=u^2*inte^(-2u)du-int((2u)*inte^(-2u)du)du`

`=u^2(e^(-2u)/(-2))-int(2u)(e^(-2u)/(-2))du`

`=-u^2/(2e^(2u))+int(u*e^(-2u))du`

Again `"using "color(blue)"Integration by Parts :"`

`I=-u^2/(2e^(2u))+[u*(e^(-2u)/(-2))-int1*(e^(-2u)/(-2))du]`

`=-u^2/(2e^(2u))-u/(2e^(2u))+1/2inte^(-2u)du`

`=-u^2/(2e^(2u))-u/(2e^(2u))+1/2*(e^(-2u)/(-2))+c`

`=-u^2/(2e^(2u))-u/(2e^(2u))-1/(4e^(2u))+c`

`=-1/(4e^(2u))[2u^2+2u+1]+c`

Subst. back , `u=lnx ande^u=x`, we get

`I=-1/(4x^2)[2(lnx)^2+2lnx+1]+c`

NOTE :

`inte^(-2u)du` , we put , `-2u=t=>u=-t/2=>du=-1/2dt`

So,

`inte^(-2u)du=inte^t(-1/2)dt=-1/2inte^tdt=-1/2e^t=e^(-2u)/(-2)`