What is the integral of (4x^2-3x+17)/((3x-1)(x^2+4)) ?

The integral of (4x^2-3x+17)/((3x-1)(x^2+4)) is 4/3 ln|3x-1|-1/2 arctan(x/2)+C

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integral of (4x^2-3x+17)/((3x-1)(x^2+4))

\int \frac{4 x ^{2} - 3 x + 17}{( 3 x - 1 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{4 x ^{2} - 3 x + 17}{( 3 x - 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{4 x ^{2} - 3 x + 17}{( 3 x - 1 ) ( x ^{2} + 4 )} : \frac{4}{3 x - 1} - \frac{1}{x ^{2} + 4}

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

\int \frac{4}{3 x - 1} d x = \frac{4}{3} ln ∣ 3 x - 1 ∣

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

= \frac{4}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

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

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What is the integral of (4x^2-3x+17)/((3x-1)(x^2+4)) ?
The integral of (4x^2-3x+17)/((3x-1)(x^2+4)) is 4/3 ln|3x-1|-1/2 arctan(x/2)+C