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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{3 x}{x ^{2} - 4 x + 20} d x = \frac{3}{2} (ln \frac{1}{16} (x - 2)^{2} + 1 + arctan (\frac{1}{4} (x - 2)))

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

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

Simplify

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

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

Add a constant to the solution

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

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