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

The integral of (x^2+x-2)/((3x-1)(x^2+1)) is -7/15 ln|3x-1|+1/5 (2ln|x^2+1|+3arctan(x))+C

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

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

- \frac{7}{15} ln ∣ 3 x - 1 ∣ + \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + x - 2}{( 3 x - 1 ) ( x ^{2} + 1 )} : - \frac{7}{5 ( 3 x - 1 )} + \frac{4 x + 3}{5 ( x ^{2} + 1 )}

= \int - \frac{7}{5 ( 3 x - 1 )} + \frac{4 x + 3}{5 ( 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{7}{5 ( 3 x - 1 )} d x + \int \frac{4 x + 3}{5 ( x ^{2} + 1 )} d x

\int \frac{7}{5 ( 3 x - 1 )} d x = \frac{7}{15} ln ∣ 3 x - 1 ∣

\int \frac{4 x + 3}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x))

= - \frac{7}{15} ln ∣ 3 x - 1 ∣ + \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x))

Add a constant to the solution

= - \frac{7}{15} ln ∣ 3 x - 1 ∣ + \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x)) + C

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What is the integral of (x^2+x-2)/((3x-1)(x^2+1)) ?
The integral of (x^2+x-2)/((3x-1)(x^2+1)) is -7/15 ln|3x-1|+1/5 (2ln|x^2+1|+3arctan(x))+C