# Integration by substitution

## Homework Statement

dz/dx=(3x2+x)(2x^3+x^2)^2[/B]

## Homework Equations

∫(3x^2+x)(2x^3+x^2)^2 dx

## The Attempt at a Solution

I tried substituting (2x^3+x^2)
Let t= 2x^3 + x^2
dt=6x^2+2x dx
dt/dx= 6x^2+2x
I can only solve till this point . I don't have any clue how to solve it further
But how do we get 1/2 ∫u^2 du ?? I don't understand it at all

member 587159

## Homework Statement

dz/dx=(3x2+x)(2x^3+x^2)^2[/B]

## Homework Equations

∫(3x^2+x)(2x^3+x^2)^2 dx

## The Attempt at a Solution

I tried substituting (2x^3+x^2)
Let t= 2x^3 + x^2
dt=6x^2+2x dx
dt/dx= 6x^2+2x
I can only solve till this point . I don't have any clue how to solve it further
But how do we get 1/2 ∫u^2 du ?? I don't understand it at all

Let u = 2x^3 + x^2
=> du = (6x^2 + 2x)dx
=> du = 2(3x^2 + x)dx
=> du/2 = (3x^2 + x)dx

Thus, the integral becomes what you need.

• Nanu Nana
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

dz/dx=(3x2+x)(2x^3+x^2)^2[/B]

## Homework Equations

∫(3x^2+x)(2x^3+x^2)^2 dx

## The Attempt at a Solution

I tried substituting (2x^3+x^2)
Let t= 2x^3 + x^2
dt=6x^2+2x dx
dt/dx= 6x^2+2x
I can only solve till this point . I don't have any clue how to solve it further
But how do we get 1/2 ∫u^2 du ?? I don't understand it at all
Well, you used t as a variable rather than u, but that's unimportant.

Looking at Math_QED's post, the main thing your solution is lacking is that you should factor a 2 out of the right hand side of ##\ dt=(6x^2+2x) dx \ .##

• Nanu Nana
Thank you both . I think i understand it now :)