Question

What is `int (5)/(1+3x)^3dx`?

Answer 1

`-5/{6(1+3x)^2}+C`

Explanation:

The integral `I=int5/(1+3x)^3dx`, can be directly dealt with, if we

use the following Rule :

`intf(x)dx=F(x)+crArrintf(ax+b)dx=1/aF(ax+b)+c', (ane0).`

`f(x)=1/x^3=x^-3," we have, "F(x)=x^(-3+1)/(-3+1)=-1/(2x^2)`.

`:. int5/(1+3x)^3dx=5int1/(1+3x)^3dx`,

`=5[1/3{-1/2*1/(1+3x)^2}]`.

`rArr I=-5/6*1/(1+3x)^2+C`.

OTHERWISE,

Let `1+3x=t," so that, "3dx=dt, or, dx=1/3dt`.

`:. I=int5/(1+3x)^3dx=5int1/(1+3x)^3dx`,

`=5int1/t^3*1/3dt`,

`=5/3intt^-3dt`,

`=5/3*t^(-3+1)/(-3+1)`,

`=-5/6*t^-2,`

`=-5/(6t^2)`.

`rArr I=-5/{6(1+3x)^2}+C`, as before!

Enjoy Maths.!