What is the integral of ((x-5))/(x^2-18x+82) ?

The integral of ((x-5))/(x^2-18x+82) is 1/2 ln|x^2-18x+82|+4arctan(x-9)+C

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integral of ((x-5))/(x^2-18x+82)

\int \frac{( x - 5 )}{x ^{2} - 18 x + 82} d x

\frac{1}{2} ln x^{2} - 18 x + 82 + 4 arctan (x - 9) + C

Solution steps

\int \frac{x - 5}{x ^{2} - 18 x + 82} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 8 u + 17} d u

= \int \frac{u}{u ^{2} - 8 u + 17} d u

Complete the square

u^{2} - 8 u + 17 : (u - 4)^{2} + 1

= \int \frac{u}{( u - 4 ) ^{2} + 1} d u

Apply u-substitution

: \int \frac{v + 4}{v ^{2} + 1} d v

= \int \frac{v + 4}{v ^{2} + 1} d v

Expand

\frac{v + 4}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{4}{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{4}{v ^{2} + 1} d v

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

\int \frac{4}{v ^{2} + 1} d v = 4 arctan (v)

= \frac{1}{2} ln v^{2} + 1 + 4 arctan (v)

Substitute back

= \frac{1}{2} ln (x - 5 - 4)^{2} + 1 + 4 arctan (x - 5 - 4)

Simplify

\frac{1}{2} ln (x - 5 - 4)^{2} + 1 + 4 arctan (x - 5 - 4) : \frac{1}{2} ln x^{2} - 18 x + 82 + 4 arctan (x - 9)

= \frac{1}{2} ln x^{2} - 18 x + 82 + 4 arctan (x - 9)

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 18 x + 82 + 4 arctan (x - 9) + C

f (x) = 1.5 x^{2} + 2 x^{1.5}

n = 1 \sum \infty \frac{1}{n ^{4} + n ^{2}}

\frac{d}{d x} (\frac{x ^{2} - 3 x - 4}{x + 1})

x \to \infty lim (\frac{3}{2 x})

\frac{d}{d x} (a x + \frac{a}{a x})

What is the integral of ((x-5))/(x^2-18x+82) ?
The integral of ((x-5))/(x^2-18x+82) is 1/2 ln|x^2-18x+82|+4arctan(x-9)+C