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

The integral of (2x^3)/(2x^4+x^2-1) is 2(1/12 ln|sqrt(2)x+1|+1/12 ln|sqrt(2)x-1|+1/6 ln|x^2+1|)+C

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integral of (2x^3)/(2x^4+x^2-1)

\int \frac{2 x ^{3}}{2 x ^{4} + x ^{2} - 1} d x

2 (\frac{1}{12} ln 2 x + 1 + \frac{1}{12} ln 2 x - 1 + \frac{1}{6} ln x^{2} + 1) + C

Solution steps

\int \frac{2 x ^{3}}{2 x ^{4} + x ^{2} - 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 ^{3}}{2 x ^{4} + x ^{2} - 1} d x

Take the partial fraction of

\frac{x ^{3}}{2 x ^{4} + x ^{2} - 1} : \frac{1}{6 2 ( 2 x + 1 )} + \frac{1}{6 2 ( 2 x - 1 )} + \frac{x}{3 ( x ^{2} + 1 )}

= 2 \cdot \int \frac{1}{6 2 ( 2 x + 1 )} + \frac{1}{6 2 ( 2 x - 1 )} + \frac{x}{3 ( 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}{6 2 ( 2 x + 1 )} d x + \int \frac{1}{6 2 ( 2 x - 1 )} d x + \int \frac{x}{3 ( x ^{2} + 1 )} d x)

\int \frac{1}{6 2 ( 2 x + 1 )} d x = \frac{1}{12} ln 2 x + 1

\int \frac{1}{6 2 ( 2 x - 1 )} d x = \frac{1}{12} ln 2 x - 1

\int \frac{x}{3 ( x ^{2} + 1 )} d x = \frac{1}{6} ln x^{2} + 1

= 2 (\frac{1}{12} ln 2 x + 1 + \frac{1}{12} ln 2 x - 1 + \frac{1}{6} ln x^{2} + 1)

Add a constant to the solution

= 2 (\frac{1}{12} ln 2 x + 1 + \frac{1}{12} ln 2 x - 1 + \frac{1}{6} ln x^{2} + 1) + C

What is the integral of (2x^3)/(2x^4+x^2-1) ?
The integral of (2x^3)/(2x^4+x^2-1) is 2(1/12 ln|sqrt(2)x+1|+1/12 ln|sqrt(2)x-1|+1/6 ln|x^2+1|)+C