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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

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

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

Simplify

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

Add a constant to the solution

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

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

f (x) = x + 2, at (x = 4)

\int 2 y e^{y^{2}} d y

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

\frac{d y}{d x} + y \cdot 1 + x^{2} = 0

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