What is the integral of (6x^2-3x+5)/((2x-1)(x^2+1)) ?

The integral of (6x^2-3x+5)/((2x-1)(x^2+1)) is 2ln|2x-1|+1/2 ln|x^2+1|-arctan(x)+C

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integral of (6x^2-3x+5)/((2x-1)(x^2+1))

\int \frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} : \frac{4}{2 x - 1} + \frac{x - 1}{x ^{2} + 1}

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

\int \frac{4}{2 x - 1} d x = 2 ln ∣ 2 x - 1 ∣

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

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

Add a constant to the solution

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

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What is the integral of (6x^2-3x+5)/((2x-1)(x^2+1)) ?
The integral of (6x^2-3x+5)/((2x-1)(x^2+1)) is 2ln|2x-1|+1/2 ln|x^2+1|-arctan(x)+C