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

The integral of (2/(y+1))^2 is (4y)/((y+1)^2)+8(-1/(y+1)+1/(2(y+1)^2))+C

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

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

\frac{4 y}{( y + 1 ) ^{2}} + 8 (- \frac{1}{y + 1} + \frac{1}{2 ( y + 1 ) ^{2}}) + C

Solution steps

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

Apply Integration By Parts

: \frac{4 y}{( y + 1 ) ^{2}} - \int - \frac{8 y}{( y + 1 ) ^{3}} d y

= \frac{4 y}{( y + 1 ) ^{2}} - \int - \frac{8 y}{( y + 1 ) ^{3}} d y

\int - \frac{8 y}{( y + 1 ) ^{3}} d y = - 8 (- \frac{1}{y + 1} + \frac{1}{2 ( y + 1 ) ^{2}})

= \frac{4 y}{( y + 1 ) ^{2}} - (- 8 (- \frac{1}{y + 1} + \frac{1}{2 ( y + 1 ) ^{2}}))

Simplify

= \frac{4 y}{( y + 1 ) ^{2}} + 8 (- \frac{1}{y + 1} + \frac{1}{2 ( y + 1 ) ^{2}})

Add a constant to the solution

= \frac{4 y}{( y + 1 ) ^{2}} + 8 (- \frac{1}{y + 1} + \frac{1}{2 ( y + 1 ) ^{2}}) + C

f (x) = x^{2} - 8, (- 2, - 4)

\frac{4}{( s ^{4} + 4 )}

\int_{2}^{\infty} x e^{- 9 x} d x

x \to 4 lim (4 - x)

\frac{1}{x - 7}

What is the integral of (2/(y+1))^2 ?
The integral of (2/(y+1))^2 is (4y)/((y+1)^2)+8(-1/(y+1)+1/(2(y+1)^2))+C