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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

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