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

The integral of (2y^2)/((x+y)^2) is 2(-2x(ln|x+y|+x/(x+y))+(x^2)/(x+y)+y)+C

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

integral

\frac{2 y ^{2}}{( x + y ) ^{2}}

2 (- 2 x (ln ∣ x + y ∣ + \frac{x}{x + y}) + \frac{x ^{2}}{x + y} + y) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{y ^{2}}{( x + y ) ^{2}} d y

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{- 2 x y - x ^{2}}{( x + y ) ^{2}} d y = - 2 x (ln ∣ x + y ∣ + \frac{x}{x + y}) + \frac{x ^{2}}{x + y}

\int 1 d y = y

= 2 (- 2 x (ln ∣ x + y ∣ + \frac{x}{x + y}) + \frac{x ^{2}}{x + y} + y)

Add a constant to the solution

= 2 (- 2 x (ln ∣ x + y ∣ + \frac{x}{x + y}) + \frac{x ^{2}}{x + y} + y) + C

\frac{1}{e ^{3}} x + sin (x)

(2 - 3 x^{2})^{3}

\frac{1}{( x + 2 )}

cos^{- 2} (x)

1 - cos (6 x)

What is the integral of (2y^2)/((x+y)^2) ?
The integral of (2y^2)/((x+y)^2) is 2(-2x(ln|x+y|+x/(x+y))+(x^2)/(x+y)+y)+C