Question

integrate the function [x²/(2 + 3x³)³].dx

Answer 1

HELLO DEAR,

given function is $\bold{\int{\frac{x^2}{(2 + 3x^3)^3}}\,dx}$

let (2 + 3x³) = t $\bold{\Rightarrow dt/dx = 9x^2}$
[te]\bold{dx = dt/9x^2}[/tex]

now,

$\bold{\int{\frac{x^2}{(2 + 3x^3)^3}*dt/9x^2}}$

$\bold{\frac{1}{9}\int{\frac{dt}{t^3}}$

$\bold{\frac{1}{9}(-\frac{1}{2t^2}) + C}$

$\bold{-\frac{1}{9(2 + 3x^3)} + C}$

where, C is the arbitrary constant.

hence, the integral of [x²/(2 + 3x³)³] = -1/9(2 + 3x³) + C.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hope it will help you!!