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

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

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

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

- \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= - \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )}

Add a constant to the solution

= - \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + C

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