What is the integral of (2x)/((x^2+1)(x^2+3)) ?

The integral of (2x)/((x^2+1)(x^2+3)) is 2(1/4 ln|x^2+1|-1/4 ln|x^2+3|)+C

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integral of (2x)/((x^2+1)(x^2+3))

\int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} d x

2 (\frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} + 3) + C

Solution steps

\int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} 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 ^{2} + 3 )} d x

Take the partial fraction of

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} : \frac{x}{2 ( x ^{2} + 1 )} - \frac{x}{2 ( x ^{2} + 3 )}

= 2 \cdot \int \frac{x}{2 ( x ^{2} + 1 )} - \frac{x}{2 ( x ^{2} + 3 )} 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}{2 ( x ^{2} + 1 )} d x - \int \frac{x}{2 ( x ^{2} + 3 )} d x)

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

\int \frac{x}{2 ( x ^{2} + 3 )} d x = \frac{1}{4} ln x^{2} + 3

= 2 (\frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} + 3)

Add a constant to the solution

= 2 (\frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} + 3) + C

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What is the integral of (2x)/((x^2+1)(x^2+3)) ?
The integral of (2x)/((x^2+1)(x^2+3)) is 2(1/4 ln|x^2+1|-1/4 ln|x^2+3|)+C