1. ## Integral Rp

I need solutions ...

$\displaystyle \int_0^2 \frac{x}{(x-1)^3}dx$

$\displaystyle \int_0^2 \frac{1}{\sqrt{x(2-x)}}dx$

2. Originally Posted by gilyos
I need solutions
And I need effort from you, unless that these are "proposed problems."

3. Originally Posted by gilyos
I need solutions ...

$\displaystyle \int_0^2 \frac{x}{(x-1)^3}dx$

$\displaystyle \int_0^2 \frac{1}{\sqrt{x(2-x)}}dx$
Where are you getting stuck?

4. Originally Posted by gilyos
I need solutions ...

$\displaystyle \int_0^2 \frac{x}{(x-1)^3}dx$

$\displaystyle \int_0^2 \frac{1}{\sqrt{x(2-x)}}dx$
1. Expand the denominator and then divide each term in the denominator by $\displaystyle x$ to simplify. Else use Integration by Parts with $\displaystyle u = x$ and $\displaystyle dv = (x - 1)^{-3}$.

2. Expand the denominator, Complete the Square and use a Trigonometric Substitution.

5. u can solve the secound Problem ?

6. Originally Posted by gilyos
u can solve the secound Problem ?
$\displaystyle x(2-x)=2x-x^2=-(x^2-2x)=-(x^2-2x-1+1)=-(x^2-2x+1)+1=1-(x-1)^2$

So can you take $\displaystyle \int\frac{1}{\sqrt{1-(x-1)^2}}dx$?

7. Originally Posted by gilyos
u can solve the secound Problem ?
You still have not shown any work. Try to follow some of the hints you were given.

For the first one, I'd use substitution, $\displaystyle u = x - 1$

For the second, substitution $\displaystyle u = \sqrt x$. you will get it in a form where you can use $\displaystyle \int \frac {1}{\sqrt{a^2 - x^2}}~dx = \sin^{-1} \frac xa + C$ to finish off the problem