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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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