Indefinite integrals

• Jan 9th 2010, 05:12 AM
mj.alawami
Indefinite integrals
Question:
$\displaystyle \int(x^2-4x+4)^\frac{4}{3} . dx$

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

But i dont know how to solve it.
• Jan 9th 2010, 05:20 AM
Defunkt
Quote:

Originally Posted by mj.alawami
Question:
$\displaystyle \int(x^2-4x+4)^\frac{4}{3} . dx$

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

But i dont know how to solve it.

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

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

Can you solve it now?
• Jan 9th 2010, 05:22 AM
Prove It
Quote:

Originally Posted by mj.alawami
Question:
$\displaystyle \int(x^2-4x+4)^\frac{4}{3} . dx$

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

But i dont know how to solve it.

You are DEFINITELY on the right track.

Now use a $\displaystyle u$ substitution.

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

The integral becomes

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

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

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

$\displaystyle = \frac{3}{11}(x - 2)^{\frac{11}{3}} + C$.
• Jan 9th 2010, 05:24 AM
mj.alawami
Quote:

Originally Posted by Defunkt
Let $\displaystyle u = (x-2)^2 \Rightarrow dx = du$.

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

Can you solve it now?

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

Is this the answer?
• Jan 9th 2010, 05:31 AM
Prove It
Quote:

Originally Posted by mj.alawami
Attempt:
$\displaystyle \frac{3}{11} (x-2)^\frac{11}{3}+C$

Is this the answer?

Yes, Well done!