What is the integral of (4pix+4pi)/(x^2-4x+8) ?

The integral of (4pix+4pi)/(x^2-4x+8) is 2piln| 1/4 x^2-x+2|+6piarctan(1/2 x-1)+C

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integral of (4pix+4pi)/(x^2-4x+8)

\int \frac{4 π x + 4 π}{x ^{2} - 4 x + 8} d x

2 π ln \frac{1}{4} x^{2} - x + 2 + 6 π arctan (\frac{1}{2} x - 1) + C

Solution steps

\int \frac{4 π x + 4 π}{x ^{2} - 4 x + 8} d x

Expand

\frac{4 π x + 4 π}{x ^{2} - 4 x + 8} : \frac{4 π x}{x ^{2} - 4 x + 8} + \frac{4 π}{x ^{2} - 4 x + 8}

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 π x}{x ^{2} - 4 x + 8} d x + \int \frac{4 π}{x ^{2} - 4 x + 8} d x

\int \frac{4 π x}{x ^{2} - 4 x + 8} d x = 4 π (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2)))

\int \frac{4 π}{x ^{2} - 4 x + 8} d x = 2 π arctan (\frac{1}{2} (x - 2))

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

Simplify

4 π (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2))) + 2 π arctan (\frac{1}{2} (x - 2)) : 2 π ln \frac{1}{4} x^{2} - x + 2 + 6 π arctan (\frac{1}{2} x - 1)

= 2 π ln \frac{1}{4} x^{2} - x + 2 + 6 π arctan (\frac{1}{2} x - 1)

Add a constant to the solution

= 2 π ln \frac{1}{4} x^{2} - x + 2 + 6 π arctan (\frac{1}{2} x - 1) + C

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What is the integral of (4pix+4pi)/(x^2-4x+8) ?
The integral of (4pix+4pi)/(x^2-4x+8) is 2piln| 1/4 x^2-x+2|+6piarctan(1/2 x-1)+C