1. ## Calculus Integration

integration (0-3) $4x/((x-5)^2)$
integration (0-inf) $x^2/(4+x^6)$

2. Originally Posted by Sally_Math
integration (0-3) $4x/((x-5)^2)$
integration (0-inf) $x^2/(4+x^6)$
For the first, use $u=x-5$.

For the second, do you think you can make soemthing if you know that $x^6=(x^3)^2$ ?
Do not forget that the second one is an improper integral.

3. i tried the u substituition but it didn't work

4. Originally Posted by Sally_Math
i tried the u substituition but it didn't work
No. It works.

5. Originally Posted by Sally_Math
i tried the u substituition but it didn't work
It's not as "obvious":

$\int_0^3 \frac{4x}{(x-5)^2}dx$

$u=x-5 \rightarrow u+5=x$

$du=dx$

$\int_0^3 \frac{4(u+5)}{u^2}du$

Can probably take it from there. . .

For the second one, gosh - that one COULD look like the derivative of a trig function if we could only do some hand-waving (re: u-sub) to make it so. . .

$\int_0^\infty \frac{x^2}{4+x^6}dx$

Let me take $u$ to be $x^3$:

$u = x^3$

$du = 3x^2dx \Rightarrow \frac{du}{3} = x^2 dx$

And well hot damn, I have an "x-squared dx". Now replace all that and see what juicy derivative you get. From here it should be "elementary. . ." - hopefully. Assuming you know your trig derivatives.

6. Originally Posted by ANDS!
It's not as "obvious":

$\int_0^3 \frac{4x}{(x-5)^2}dx$

$u=x-5 \rightarrow u+5=x$

$du=dx$

$\int_0^3 \frac{4(u+5)}{u^2}du$

Can probably take it from there. . .
You should change the upper/lower limits.

7. Well look at that I should. . .I'm actually going to say I left them there as a learning tool for the OP to see if he can catch mistakes - yup. That's what I did.