What is the integral of (x+4)/(x^2+6x+18) ?

The integral of (x+4)/(x^2+6x+18) is 1/2 ln|x^2+6x+18|+1/3 arctan((x+3)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+4)/(x^2+6x+18)

\int \frac{x + 4}{x ^{2} + 6 x + 18} d x

\frac{1}{2} ln x^{2} + 6 x + 18 + \frac{1}{3} arctan (\frac{x + 3}{3}) + C

Solution steps

\int \frac{x + 4}{x ^{2} + 6 x + 18} d x

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{v + 1}{v ^{2} + 9} d v

= \int \frac{v + 1}{v ^{2} + 9} d v

Expand

\frac{v + 1}{v ^{2} + 9} : \frac{v}{v ^{2} + 9} + \frac{1}{v ^{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{v}{v ^{2} + 9} d v + \int \frac{1}{v ^{2} + 9} d v

\int \frac{v}{v ^{2} + 9} d v = \frac{1}{2} ln v^{2} + 9

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

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

Substitute back

= \frac{1}{2} ln (x + 4 - 1)^{2} + 9 + \frac{1}{3} arctan (\frac{x + 4 - 1}{3})

Simplify

\frac{1}{2} ln (x + 4 - 1)^{2} + 9 + \frac{1}{3} arctan (\frac{x + 4 - 1}{3}) : \frac{1}{2} ln x^{2} + 6 x + 18 + \frac{1}{3} arctan (\frac{x + 3}{3})

= \frac{1}{2} ln x^{2} + 6 x + 18 + \frac{1}{3} arctan (\frac{x + 3}{3})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 18 + \frac{1}{3} arctan (\frac{x + 3}{3}) + C

n = 3 \sum \infty (\frac{4}{10})^{n}

\frac{d}{d x} (4 (5 - 8 x)^{5})

\frac{\partial}{\partial x} (x^{2} y + 10 x y)

y = ln (8) x^{2}

(2 x - 9)^{2}

What is the integral of (x+4)/(x^2+6x+18) ?
The integral of (x+4)/(x^2+6x+18) is 1/2 ln|x^2+6x+18|+1/3 arctan((x+3)/3)+C