What is the integral of (x-7)/(x^2-18x+82) ?

The integral of (x-7)/(x^2-18x+82) is 1/2 ln|x^2-18x+82|+2arctan(x-9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x-7)/(x^2-18x+82)

\int \frac{x - 7}{x ^{2} - 18 x + 82} d x

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

Solution steps

\int \frac{x - 7}{x ^{2} - 18 x + 82} d x

Apply u-substitution

= \int \frac{u}{u ^{2} - 4 u + 5} d u

Complete the square

u^{2} - 4 u + 5 : (u - 2)^{2} + 1

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

Apply u-substitution

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

Expand

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

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

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

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

Substitute back

= \frac{1}{2} ln (x - 7 - 2)^{2} + 1 + 2 arctan (x - 7 - 2)

Simplify

\frac{1}{2} ln (x - 7 - 2)^{2} + 1 + 2 arctan (x - 7 - 2) : \frac{1}{2} ln x^{2} - 18 x + 82 + 2 arctan (x - 9)

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

Add a constant to the solution

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

What is the integral of (x-7)/(x^2-18x+82) ?
The integral of (x-7)/(x^2-18x+82) is 1/2 ln|x^2-18x+82|+2arctan(x-9)+C