What is the integral of (2x^2+3x+18)/((2x+1)(x^2+4)) ?

The integral of (2x^2+3x+18)/((2x+1)(x^2+4)) is 2ln|2x+1|-1/2 ln|x^2+4|+arctan(x/2)+C

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integral of (2x^2+3x+18)/((2x+1)(x^2+4))

\int \frac{2 x ^{2} + 3 x + 18}{( 2 x + 1 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{2 x ^{2} + 3 x + 18}{( 2 x + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{2 x ^{2} + 3 x + 18}{( 2 x + 1 ) ( x ^{2} + 4 )} : \frac{4}{2 x + 1} + \frac{- x + 2}{x ^{2} + 4}

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

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

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

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

Add a constant to the solution

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

What is the integral of (2x^2+3x+18)/((2x+1)(x^2+4)) ?
The integral of (2x^2+3x+18)/((2x+1)(x^2+4)) is 2ln|2x+1|-1/2 ln|x^2+4|+arctan(x/2)+C