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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{( u - 1 ) u ^{2}} d u

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

Take the partial fraction of

\frac{1}{( u - 1 ) u ^{2}} : \frac{1}{u - 1} - \frac{1}{u} - \frac{1}{u ^{2}}

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

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}{u - 1} d u - \int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u - 1 ∣ - ln ∣ u ∣ - (- \frac{1}{u})

Substitute back

u = 1 + x

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

Simplify

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

Add a constant to the solution

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

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