What is the integral of 1/(x^2(x+1)^3) ?

The integral of 1/(x^2(x+1)^3) is -3ln|x|-1/x+3ln|x+1|-2/(x+1)-1/(2(x+1)^2)+C

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integral of 1/(x^2(x+1)^3)

\int \frac{1}{x ^{2} ( x + 1 ) ^{3}} d x

- 3 ln ∣ x ∣ - \frac{1}{x} + 3 ln ∣ x + 1 ∣ - \frac{2}{x + 1} - \frac{1}{2 ( x + 1 ) ^{2}} + C

Solution steps

\int \frac{1}{x ^{2} ( x + 1 ) ^{3}} d x

Take the partial fraction of

\frac{1}{x ^{2} ( x + 1 ) ^{3}} : - \frac{3}{x} + \frac{1}{x ^{2}} + \frac{3}{x + 1} + \frac{2}{( x + 1 ) ^{2}} + \frac{1}{( x + 1 ) ^{3}}

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

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

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

= - 3 ln ∣ x ∣ - \frac{1}{x} + 3 ln ∣ x + 1 ∣ - \frac{2}{x + 1} - \frac{1}{2 ( x + 1 ) ^{2}}

Add a constant to the solution

= - 3 ln ∣ x ∣ - \frac{1}{x} + 3 ln ∣ x + 1 ∣ - \frac{2}{x + 1} - \frac{1}{2 ( x + 1 ) ^{2}} + C

f (x) = \frac{5 x}{x - 3}

n = 1 \sum \infty (2)^{n - 1}

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

\int \frac{4}{x ^{2} - 1} d x

\frac{d}{d x} (- \frac{1}{2} sin (x))

What is the integral of 1/(x^2(x+1)^3) ?
The integral of 1/(x^2(x+1)^3) is -3ln|x|-1/x+3ln|x+1|-2/(x+1)-1/(2(x+1)^2)+C