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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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