What is the integral of (x-6)/(x^2-18x+82) ?

The integral of (x-6)/(x^2-18x+82) is 1/2 ln|x^2-18x+82|+3arctan(x-9)+C

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integral of (x-6)/(x^2-18x+82)

\int \frac{x - 6}{x ^{2} - 18 x + 82} d x

\frac{1}{2} ln x^{2} - 18 x + 82 + 3 arctan (x - 9) + C

Solution steps

\int \frac{x - 6}{x ^{2} - 18 x + 82} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 6 u + 10} d u

= \int \frac{u}{u ^{2} - 6 u + 10} d u

Complete the square

u^{2} - 6 u + 10 : (u - 3)^{2} + 1

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

Apply u-substitution

: \int \frac{v + 3}{v ^{2} + 1} d v

= \int \frac{v + 3}{v ^{2} + 1} d v

Expand

\frac{v + 3}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{3}{v ^{2} + 1}

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{v}{v ^{2} + 1} d v + \int \frac{3}{v ^{2} + 1} d v

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

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= \frac{1}{2} ln v^{2} + 1 + 3 arctan (v)

Substitute back

= \frac{1}{2} ln (x - 6 - 3)^{2} + 1 + 3 arctan (x - 6 - 3)

Simplify

\frac{1}{2} ln (x - 6 - 3)^{2} + 1 + 3 arctan (x - 6 - 3) : \frac{1}{2} ln x^{2} - 18 x + 82 + 3 arctan (x - 9)

= \frac{1}{2} ln x^{2} - 18 x + 82 + 3 arctan (x - 9)

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 18 x + 82 + 3 arctan (x - 9) + C

\int t^{2} e^{- 17 t} d t

x \to 3 lim (9 x + 92)

\frac{d}{d x} ((f (x))^{n})

\int e^{- 18 x} d x

\frac{\partial}{\partial t} (s - t)

What is the integral of (x-6)/(x^2-18x+82) ?
The integral of (x-6)/(x^2-18x+82) is 1/2 ln|x^2-18x+82|+3arctan(x-9)+C