Indefinite integrals

**mj.alawami** · January 9th 2010, 05:12 AM

Question:
$\int(x^2-4x+4)^\frac{4}{3} . dx$

i know that $x^2-4x+4$ simplifies to $(x-2)^2$
So you can say the equation is:
$\int (x-2)^\frac{8}{3} .dx$

But i dont know how to solve it.
Please answer in steps Thank you

**Defunkt** · January 9th 2010, 05:20 AM

Originally Posted by mj.alawami

Question:
$\int(x^2-4x+4)^\frac{4}{3} . dx$

i know that $x^2-4x+4$ simplifies to $(x-2)^2$
So you can say the equation is:
$\int (x-2)^\frac{8}{3} .dx$

But i dont know how to solve it.
Please answer in steps Thank you

Let $u = x-2 \Rightarrow dx = du$ .

Then: $\int (x-2)^\frac{8}{3} ~dx = \int u^{\frac{8}{3}} ~ du$

Can you solve it now?

**Prove It** · January 9th 2010, 05:22 AM

Originally Posted by mj.alawami

Question:
$\int(x^2-4x+4)^\frac{4}{3} . dx$

i know that $x^2-4x+4$ simplifies to $(x-2)^2$
So you can say the equation is:
$\int (x-2)^\frac{8}{3} .dx$

But i dont know how to solve it.
Please answer in steps Thank you

You are DEFINITELY on the right track.

Now use a $u$ substitution.

Let $u = x - 2$ so that $\frac{du}{dx} = 1$ .

The integral becomes

$\int{u^{\frac{8}{3}}\,\frac{du}{dx}\,dx}$

$= \int{u^{\frac{8}{3}}\,du}$

$= \frac{3}{11}u^{\frac{11}{3}} + C$

$= \frac{3}{11}(x - 2)^{\frac{11}{3}} + C$ .

**mj.alawami** · January 9th 2010, 05:24 AM

Originally Posted by Defunkt

Let $u = (x-2)^2 \Rightarrow dx = du$ .

Then: $\int (x-2)^\frac{8}{3} ~dx = \int u^{\frac{8}{3}} ~ du$

Can you solve it now?

Attempt:
$\frac{3}{11} (x-2)^\frac{11}{3}+C$

Is this the answer?

**Prove It** · January 9th 2010, 05:31 AM

Originally Posted by mj.alawami

Attempt:
$\frac{3}{11} (x-2)^\frac{11}{3}+C$

Is this the answer?

Yes, Well done!

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