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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int \frac{a + x}{a - x} d x

f (x) = \frac{1}{3 - x}

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

x \to 0 lim ((1 - 2 x)^{\frac{2}{x}})

\int \frac{4 x ^{2} - 21 x}{( x - 3 ) ^{2} ( 2 x + 3 )} d x

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