# 10 POINTS! Find the given integral and check your answer by differentiation. y^2/(y^3 + 4)^2 dy?

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Find `int y^2/((y^3+4)^2)dy` :

Let `u=y^3+4,du=3y^2dy`

Then `int yy^2/((y^3+4)^2)dy=1/3int(3y^2)/((y^3+4)^2)dy`

`=1/3intu^(-2)du`

Using the general power rule we can integrate:

`=1/3(-1)u^(-1)+C`

`=-1/3u^(-1)+C` substituting for u we get:

`=-1/3(y^3+4)^(-1)+C` or

`=-1/(3(y^3+4))+C`

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Check: `d/(dy) -1/(3(y^3+4))+C=d/(dy) -1/3(y^3+4)^(-1)+C`

`=(-1/3)(-1)(y^3+4)^(-2)(3y^2)`

`=y^2/((y^3+4)^2)` as required.

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