What is the integral of (y^2)/((y+1)^4) ?

The integral of (y^2)/((y+1)^4) is -(y^2)/(3(y+1)^3)+(-2y-1)/(3(y+1)^2)+C

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integral of (y^2)/((y+1)^4)

\int \frac{y ^{2}}{( y + 1 ) ^{4}} d y

- \frac{y ^{2}}{3 ( y + 1 ) ^{3}} + \frac{- 2 y - 1}{3 ( y + 1 ) ^{2}} + C

Solution steps

\int \frac{y ^{2}}{( y + 1 ) ^{4}} d y

Apply Integration By Parts

: - \frac{y ^{2}}{3 ( y + 1 ) ^{3}} - \int - \frac{2 y}{3 ( y + 1 ) ^{3}} d y

= - \frac{y ^{2}}{3 ( y + 1 ) ^{3}} - \int - \frac{2 y}{3 ( y + 1 ) ^{3}} d y

\int - \frac{2 y}{3 ( y + 1 ) ^{3}} d y = - \frac{- 2 y - 1}{3 ( y + 1 ) ^{2}}

= - \frac{y ^{2}}{3 ( y + 1 ) ^{3}} - (- \frac{- 2 y - 1}{3 ( y + 1 ) ^{2}})

Simplify

= - \frac{y ^{2}}{3 ( y + 1 ) ^{3}} + \frac{- 2 y - 1}{3 ( y + 1 ) ^{2}}

Add a constant to the solution

= - \frac{y ^{2}}{3 ( y + 1 ) ^{3}} + \frac{- 2 y - 1}{3 ( y + 1 ) ^{2}} + C

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What is the integral of (y^2)/((y+1)^4) ?
The integral of (y^2)/((y+1)^4) is -(y^2)/(3(y+1)^3)+(-2y-1)/(3(y+1)^2)+C