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

The integral of (x^2)/((x+1)(x+2)^2) is ln|x+1|+4/(x+2)+C

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

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

ln ∣ x + 1 ∣ + \frac{4}{x + 2} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2}}{( x + 1 ) ( x + 2 ) ^{2}} : \frac{1}{x + 1} - \frac{4}{( x + 2 ) ^{2}}

= \int \frac{1}{x + 1} - \frac{4}{( x + 2 ) ^{2}} d x

Apply the Sum Rule:

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

= \int \frac{1}{x + 1} d x - \int \frac{4}{( x + 2 ) ^{2}} d x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

\int \frac{4}{( x + 2 ) ^{2}} d x = - \frac{4}{x + 2}

= ln ∣ x + 1 ∣ - (- \frac{4}{x + 2})

Simplify

= ln ∣ x + 1 ∣ + \frac{4}{x + 2}

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{4}{x + 2} + C

\int x^{2} e^{x y} d y

\int e^{\frac{- x}{3}} d x

\int_{- 2}^{3} \frac{27}{x ^{4}} d x

x y^{2} y^{^{'}} = y^{3} - 1 x^{3}

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

What is the integral of (x^2)/((x+1)(x+2)^2) ?
The integral of (x^2)/((x+1)(x+2)^2) is ln|x+1|+4/(x+2)+C