Integration- is it possible to integrate this?

[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) = u^6/5

x= u^6/5 - 1

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

6/5 * integral of (u^6/5 - 1)/(u^2/5 - u^1/5) du

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

 

[D H]

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

What made you pick that??

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

 

[Zoe-b]

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.

 

[Zoe-b]

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 =/

 

[D H]

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

Hint: What is (u⁶-1)/(u-1)?

 

[Zoe-b]

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