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

The integral of (x^2+1)/((x+2)(x+1)^2) is 5ln|x+2|-4ln|x+1|-2/(x+1)+C

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

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

5 ln ∣ x + 2 ∣ - 4 ln ∣ x + 1 ∣ - \frac{2}{x + 1} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 1}{( x + 2 ) ( x + 1 ) ^{2}} : \frac{5}{x + 2} - \frac{4}{x + 1} + \frac{2}{( x + 1 ) ^{2}}

= \int \frac{5}{x + 2} - \frac{4}{x + 1} + \frac{2}{( x + 1 ) ^{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{5}{x + 2} d x - \int \frac{4}{x + 1} d x + \int \frac{2}{( x + 1 ) ^{2}} d x

\int \frac{5}{x + 2} d x = 5 ln ∣ x + 2 ∣

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

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

= 5 ln ∣ x + 2 ∣ - 4 ln ∣ x + 1 ∣ - \frac{2}{x + 1}

Add a constant to the solution

= 5 ln ∣ x + 2 ∣ - 4 ln ∣ x + 1 ∣ - \frac{2}{x + 1} + C

What is the integral of (x^2+1)/((x+2)(x+1)^2) ?
The integral of (x^2+1)/((x+2)(x+1)^2) is 5ln|x+2|-4ln|x+1|-2/(x+1)+C