1. ## substitution integration

hmmm...I have been studying some of the internet sites on this subject and now I think I have confused myself even more...this is an independent study class and the textbook shows very few examples...

integral (y^2)/(y^3+5)^2 dy

u=y^3 +5
du = 3y^2
dx = y^2?

y/u integral du/3y^2 dx

y/3y^2 ln[u] +c

y/3y^2 ln [y^3+5] + C

2. Originally Posted by startingover
hmmm...I have been studying some of the internet sites on this subject and now I think I have confused myself even more...this is an independent study class and the textbook shows very few examples...

integral (y^2)/(y^3+5)^2 dy

u=y^3 +5
du = 3y^2
dx = y^2?
$\int \frac{y^2}{(y^3+5)^2} dy$

Let $x=y^3+5 \Rightarrow x'=3y^2$

Thus,

$\frac{1}{3} \frac{1}{x^2} dx$

$-\frac{1}{3} \cdot \frac{1}{x} +C$

$-\frac{1}{3}\cdot \frac{1}{y^3+5} +C$

3. Originally Posted by startingover
hmmm...I have been studying some of the internet sites on this subject and now I think I have confused myself even more...this is an independent study class and the textbook shows very few examples...

integral (y^2)/(y^3+5)^2 dy

u=y^3 +5
du = 3y^2
dx = y^2?

y/u integral du/3y^2 dx

y/3y^2 ln[u] +c

y/3y^2 ln [y^3+5] + C
it seems you have confused yourself.

$\int \frac {y^2}{\left( y^3 + 5 \right)^2}dy$

Let $u = y^3 + 5$

$\Rightarrow du = 3y^2 dy$

$\Rightarrow \frac {1}{3}du = y^2 dy$

So our integral becomes:

$\frac {1}{3} \int \frac {1}{u^2} du = \frac {1}{3} \int u^{-2}du$

$= - \frac {1}{3} u^{-1} + C$

$= - \frac {1}{3u} + C$

$= - \frac {1}{3 \left( y^3 + 5 \right)} + C$

EDIT: You're too fast for me TPH

4. how did we get -1/3?

5. Originally Posted by startingover
how did we get -1/3?
By the power rule:

$\int x^{n} dx = \frac {x^{n+1}}{n + 1} + C$

So, $\int u^{-2}du = \frac {u^{-2 + 1}}{-2 + 1} + C = - \frac {1}{u} + C$

we had the $\frac {1}{3}$ already, so we made it negative because of the minus sign we get from the integral of $u^{-2}$