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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} = 2 y + 1

n = 0 \sum \infty \frac{2}{1.1 5 ^{n}}

y^{^{''}} + 25 y = - 30 sec (5 t)

\frac{d}{d x} (\frac{3 x ^{2} - 6}{2 x + 3})

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

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