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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{2 x + 3} d x = ln ∣ 2 x + 3 ∣

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

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

Add a constant to the solution

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

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