# Integration- is it possible to integrate this?

• Zoe-b
In summary, the problem is finding the integral of x over ((1+x)1/2-(1+x)1/3)dx using integration by substitution, with a suggested substitution of u=(1+x)1/6. After some attempts, the correct substitution is u=(1+x)1/6, leading to the integral of 6 * ((u9 - u3) / (u - 1)) du. Simplifying this leads to the final answer of (2/3 * u8) + (27/4 * u7) + (216/7 * u6) + 84u5 + (756/5 * u4) + 189u3 + 166u2 +
Zoe-b

## Homework Statement

Find the integral of x over ((1+x)1/2-(1+x)1/3) dx

## Homework Equations

integration by substitution.

## The Attempt at a Solution

I have tried the substitution u = (1+x)5/6 but not really sure if this makes it any better!
du/dx = 5/6 * (1+x)-1/6

6/5 * du = (1+x)-1/6 dx

(1+x) = u6/5

x= u6/5 - 1

Take a factor of (1+x)1/6 of the denominator and rewrite to give:

6/5 * integral of (u6/5 - 1)/(u2/5 - u1/5) du

Is is possible to integrate this? Or do I need a completely different substitution?
Thanks.

Zoe-b said:
I have tried the substitution u = (1+x)5/6 but not really sure if this makes it any better!

Try the much simpler u=(1+x)1/6 (alternatively, u6=1+x).

Hmmn I was trying to get something so that the integral du/dx was a factor. I thought I'd already tried and failed with (1+x)^1/6 but will try again! Thanks.

Ok soo here goes:
u = (1+x)1/6

du/dx = 1/6 * (1+x)-5/6

6du = (1+x)-5/6 dx

u6 = (1+x)

x = u6 -1

So integral of x over (1+x)-5/6 * ((1+x)-1/3 - (1+x)-1/2)dx

Subst for u to give:
6 * integral of (u6 - 1) / (u-2 - u-3) du

= 6 * integral of (u9 - u3) / (u - 1) du

From here I made a second substitution of z = u - 1 (du = dz) which gives
6 * integral of ((z+1)9 - (z+1)3) / z dz

Using binomial theorem I get:

(2/3 * z9) + (27/4 * z8) + (216/7 * z7) + 84z6 + (756/5 * z5) + 189z4 + 166z3 + 99z2 + 36z

where z = ((1+x)1/6 -1)

Not exactly the neatest answer =/

Rather than making another substitution, why don't you just simplify (u9-u3)/(u-1)?

Hint: What is (u6-1)/(u-1)?

Yep I'm being very slow! Thanks should be able to finish now. Thought that answer looked ridiculously long.

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