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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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