What is the integral of (3x-1)/(x^2-2x+10) ?

The integral of (3x-1)/(x^2-2x+10) is 3/2 ln|x^2-2x+10|+2/3 arctan(1/3 (x-1))+C

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integral of (3x-1)/(x^2-2x+10)

\int \frac{3 x - 1}{x ^{2} - 2 x + 10} d x

\frac{3}{2} ln x^{2} - 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x - 1)) + C

Solution steps

\int \frac{3 x - 1}{x ^{2} - 2 x + 10} d x

Complete the square

x^{2} - 2 x + 10 : (x - 1)^{2} + 9

= \int \frac{3 x - 1}{( x - 1 ) ^{2} + 9} d x

Apply u-substitution

= \int \frac{3 u + 2}{u ^{2} + 9} d u

Expand

\frac{3 u + 2}{u ^{2} + 9} : \frac{3 u}{u ^{2} + 9} + \frac{2}{u ^{2} + 9}

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 u}{u ^{2} + 9} d u + \int \frac{2}{u ^{2} + 9} d u

\int \frac{3 u}{u ^{2} + 9} d u = \frac{3}{2} ln u^{2} + 9

\int \frac{2}{u ^{2} + 9} d u = \frac{2}{3} arctan (\frac{u}{3})

= \frac{3}{2} ln u^{2} + 9 + \frac{2}{3} arctan (\frac{u}{3})

Substitute back

u = x - 1

= \frac{3}{2} ln (x - 1)^{2} + 9 + \frac{2}{3} arctan (\frac{x - 1}{3})

Simplify

\frac{3}{2} ln (x - 1)^{2} + 9 + \frac{2}{3} arctan (\frac{x - 1}{3}) : \frac{3}{2} ln x^{2} - 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x - 1))

= \frac{3}{2} ln x^{2} - 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x - 1))

Add a constant to the solution

= \frac{3}{2} ln x^{2} - 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x - 1)) + C

What is the integral of (3x-1)/(x^2-2x+10) ?
The integral of (3x-1)/(x^2-2x+10) is 3/2 ln|x^2-2x+10|+2/3 arctan(1/3 (x-1))+C