What is the integral of (x-2)/(x^2-10x+106) ?

The integral of (x-2)/(x^2-10x+106) is 1/2 ln|x^2-10x+106|+1/3 arctan(1/9 (x-5))+C

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integral of (x-2)/(x^2-10x+106)

\int \frac{x - 2}{x ^{2} - 10 x + 106} d x

\frac{1}{2} ln x^{2} - 10 x + 106 + \frac{1}{3} arctan (\frac{1}{9} (x - 5)) + C

Solution steps

\int \frac{x - 2}{x ^{2} - 10 x + 106} d x

Complete the square

x^{2} - 10 x + 106 : (x - 5)^{2} + 81

= \int \frac{x - 2}{( x - 5 ) ^{2} + 81} d x

Apply u-substitution

= \int \frac{u + 3}{u ^{2} + 81} d u

Expand

\frac{u + 3}{u ^{2} + 81} : \frac{u}{u ^{2} + 81} + \frac{3}{u ^{2} + 81}

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{u}{u ^{2} + 81} d u + \int \frac{3}{u ^{2} + 81} d u

\int \frac{u}{u ^{2} + 81} d u = \frac{1}{2} ln u^{2} + 81

\int \frac{3}{u ^{2} + 81} d u = \frac{1}{3} arctan (\frac{u}{9})

= \frac{1}{2} ln u^{2} + 81 + \frac{1}{3} arctan (\frac{u}{9})

Substitute back

u = x - 5

= \frac{1}{2} ln (x - 5)^{2} + 81 + \frac{1}{3} arctan (\frac{x - 5}{9})

Simplify

\frac{1}{2} ln (x - 5)^{2} + 81 + \frac{1}{3} arctan (\frac{x - 5}{9}) : \frac{1}{2} ln x^{2} - 10 x + 106 + \frac{1}{3} arctan (\frac{1}{9} (x - 5))

= \frac{1}{2} ln x^{2} - 10 x + 106 + \frac{1}{3} arctan (\frac{1}{9} (x - 5))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 10 x + 106 + \frac{1}{3} arctan (\frac{1}{9} (x - 5)) + C

What is the integral of (x-2)/(x^2-10x+106) ?
The integral of (x-2)/(x^2-10x+106) is 1/2 ln|x^2-10x+106|+1/3 arctan(1/9 (x-5))+C