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

The integral of x/(((x+1)^2(x+2)^2)) is 3ln|x+1|+1/(x+1)-3ln|x+2|+2/(x+2)+C

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

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

3 ln ∣ x + 1 ∣ + \frac{1}{x + 1} - 3 ln ∣ x + 2 ∣ + \frac{2}{x + 2} + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x + 1 ) ^{2} ( x + 2 ) ^{2}} : \frac{3}{x + 1} - \frac{1}{( x + 1 ) ^{2}} - \frac{3}{x + 2} - \frac{2}{( x + 2 ) ^{2}}

= \int \frac{3}{x + 1} - \frac{1}{( x + 1 ) ^{2}} - \frac{3}{x + 2} - \frac{2}{( 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{3}{x + 1} d x - \int \frac{1}{( x + 1 ) ^{2}} d x - \int \frac{3}{x + 2} d x - \int \frac{2}{( x + 2 ) ^{2}} d x

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

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

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

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

= 3 ln ∣ x + 1 ∣ - (- \frac{1}{x + 1}) - 3 ln ∣ x + 2 ∣ - (- \frac{2}{x + 2})

Simplify

= 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1} - 3 ln ∣ x + 2 ∣ + \frac{2}{x + 2}

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1} - 3 ln ∣ x + 2 ∣ + \frac{2}{x + 2} + C

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What is the integral of x/(((x+1)^2(x+2)^2)) ?
The integral of x/(((x+1)^2(x+2)^2)) is 3ln|x+1|+1/(x+1)-3ln|x+2|+2/(x+2)+C