What is the integral of (x^2-x-21)/((x^2+4)(2x-1)) ?

The integral of (x^2-x-21)/((x^2+4)(2x-1)) is 3/2 ln|x^2+4|+1/2 arctan(x/2)-5/2 ln|2x-1|+C

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integral of (x^2-x-21)/((x^2+4)(2x-1))

\int \frac{x ^{2} - x - 21}{( x ^{2} + 4 ) ( 2 x - 1 )} d x

\frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) - \frac{5}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{x ^{2} - x - 21}{( x ^{2} + 4 ) ( 2 x - 1 )} d x

Take the partial fraction of

\frac{x ^{2} - x - 21}{( x ^{2} + 4 ) ( 2 x - 1 )} : \frac{3 x + 1}{x ^{2} + 4} - \frac{5}{2 x - 1}

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

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

\int \frac{5}{2 x - 1} d x = \frac{5}{2} ln ∣ 2 x - 1 ∣

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

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) - \frac{5}{2} ln ∣ 2 x - 1 ∣ + C

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What is the integral of (x^2-x-21)/((x^2+4)(2x-1)) ?
The integral of (x^2-x-21)/((x^2+4)(2x-1)) is 3/2 ln|x^2+4|+1/2 arctan(x/2)-5/2 ln|2x-1|+C