Question

Find the following integral:

Answer 1

\[\int\limits_{}^{} y^2/ (y+1)^4\]

Answer 2

What can I use for substitution?

Answer 3

let \(u=y+1\) see what happens

Answer 4

gotta love those deceptively simple u-subs :)

Answer 5

i got -u^-1 - ((u^-3) / 3)
do I convert back to x now or is that incorrect?

Answer 6

so
u = y + 1
du = dy

thus we have the integral

\[\int\limits_{}^{}\frac{y^{2}du}{u}\]

so lets write y in terms of u

y = u - 1
so we are left with

\[\int\limits_{}^{}\frac{(u-1)^{2}du}{u} = \int\limits_{}^{}\frac{(u^2 -2u + 1d)u}{u}=\int\limits_{}^{}\frac{u(u -2 + \frac{1}{u})du}{u} =\int\limits_{}^{}u -2 + \frac{1}{u}du \]

Answer 7

now just split the integral and enjoy, learning algebra manipulation is crucial to surviving calculus :)

Answer 8

yeah but you forgot the exponent, the denominator is (y+1)^4

Answer 9

oh crud you are right my bad

Answer 10

but that doesn't make the problem any harder, or the approach any diffrerent

Answer 11

yeah I have a rough idea of how to do it now

Answer 12

yeah you just have to factor out more u :)

Answer 13

\[\int\limits_{}^{}\frac{u^{4}(\frac{1}{u^{2}} -\frac{2}{u^{3}} + \frac{1}{u^{4}})du}{u^{4}} = \int\limits_{}^{}(\frac{1}{u^{2}} -\frac{2}{u^{3}} + \frac{1}{u^{4}})du\]

Answer 14

now just split the integral and use the exponent rule, and remove the 2 from the one integral

Answer 15

I would have used partial fractions, then A/(y+1) + B/(y+1)^2 + C/(y+1)^3 + D/(y+1)^4 then solve this = y^2/(y+1)^4 , you have to multiply this out. plenty of algebra. so it you multiply out the common denominator then you can solve these one by one. this will get your values for A,B,C, and D. then you integrate an easier approach.

Answer 16

easier?
please set up that system and show me how PF is easier here

Answer 17

you would have to expand cubics and all kind of ugliness

Answer 18

personally I dont like partial fractions.

Answer 19

it could be done, but in this case it's like choosing the long painful way