What is the integral of (3x^2+1)/(x^2-4x+8) ?

The integral of (3x^2+1)/(x^2-4x+8) is 3(2ln|x^2-4x+8|+x)+1/2 arctan((x-2)/2)+C

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integral of (3x^2+1)/(x^2-4x+8)

\int \frac{3 x ^{2} + 1}{x ^{2} - 4 x + 8} d x

3 (2 ln x^{2} - 4 x + 8 + x) + \frac{1}{2} arctan (\frac{x - 2}{2}) + C

Solution steps

\int \frac{3 x ^{2} + 1}{x ^{2} - 4 x + 8} d x

Expand

\frac{3 x ^{2} + 1}{x ^{2} - 4 x + 8} : \frac{3 x ^{2}}{x ^{2} - 4 x + 8} + \frac{1}{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{3 x ^{2}}{x ^{2} - 4 x + 8} d x + \int \frac{1}{x ^{2} - 4 x + 8} d x

\int \frac{3 x ^{2}}{x ^{2} - 4 x + 8} d x = 3 (2 ln x^{2} - 4 x + 8 + x)

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

= 3 (2 ln x^{2} - 4 x + 8 + x) + \frac{1}{2} arctan (\frac{x - 2}{2})

Add a constant to the solution

= 3 (2 ln x^{2} - 4 x + 8 + x) + \frac{1}{2} arctan (\frac{x - 2}{2}) + C

What is the integral of (3x^2+1)/(x^2-4x+8) ?
The integral of (3x^2+1)/(x^2-4x+8) is 3(2ln|x^2-4x+8|+x)+1/2 arctan((x-2)/2)+C