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

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

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

integral

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

- 2 ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

= - 2 ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ - \frac{1}{x + 1}

Add a constant to the solution

= - 2 ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C

\frac{( x ^{2} )}{3}

\frac{x}{( 0.5 x - 1 ) ^{0.5}}

\frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

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

2 e^{\frac{- x}{3}}

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