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

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

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

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

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

Solution steps

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (e^{- 2 x} \cdot cos (3 x))

x \to 0 + lim (3 + x^{5} ln (x))

\frac{\partial}{\partial y} (x^{3} + y z^{2})

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

\frac{d}{d x} (csc (cot (x)))

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