Indefinite integrals

• Jan 9th 2010, 06:12 AM
mj.alawami
Indefinite integrals
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.
• Jan 9th 2010, 06:20 AM
Defunkt
Quote:

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.

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?
• Jan 9th 2010, 06:22 AM
Prove It
Quote:

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.

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

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

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

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

$= \frac{3}{11}(x - 2)^{\frac{11}{3}} + C$.
• Jan 9th 2010, 06:24 AM
mj.alawami
Quote:

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

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