What is the integral of (4x-3)/((x^2+4x+20)) ?

The integral of (4x-3)/((x^2+4x+20)) is 2ln| 1/16 (x+2)^2+1|-11/4 arctan(1/4 (x+2))+C

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integral of (4x-3)/((x^2+4x+20))

\int \frac{4 x - 3}{( x ^{2} + 4 x + 20 )} d x

2 ln \frac{1}{16} (x + 2)^{2} + 1 - \frac{11}{4} arctan (\frac{1}{4} (x + 2)) + C

Solution steps

\int \frac{4 x - 3}{x ^{2} + 4 x + 20} d x

Expand

\frac{4 x - 3}{x ^{2} + 4 x + 20} : \frac{4 x}{x ^{2} + 4 x + 20} - \frac{3}{x ^{2} + 4 x + 20}

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 x}{x ^{2} + 4 x + 20} d x - \int \frac{3}{x ^{2} + 4 x + 20} d x

\int \frac{4 x}{x ^{2} + 4 x + 20} d x = 2 (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2)))

\int \frac{3}{x ^{2} + 4 x + 20} d x = \frac{3}{4} arctan (\frac{1}{4} (x + 2))

= 2 (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2))) - \frac{3}{4} arctan (\frac{1}{4} (x + 2))

Simplify

2 (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2))) - \frac{3}{4} arctan (\frac{1}{4} (x + 2)) : 2 ln \frac{1}{16} (x + 2)^{2} + 1 - \frac{11}{4} arctan (\frac{1}{4} (x + 2))

= 2 ln \frac{1}{16} (x + 2)^{2} + 1 - \frac{11}{4} arctan (\frac{1}{4} (x + 2))

Add a constant to the solution

= 2 ln \frac{1}{16} (x + 2)^{2} + 1 - \frac{11}{4} arctan (\frac{1}{4} (x + 2)) + C

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What is the integral of (4x-3)/((x^2+4x+20)) ?
The integral of (4x-3)/((x^2+4x+20)) is 2ln| 1/16 (x+2)^2+1|-11/4 arctan(1/4 (x+2))+C