What is the integral of x/(x^2+8x+20) ?

The integral of x/(x^2+8x+20) is 1/2 ln| 1/4 (x+4)^2+1|-2arctan(1/2 (x+4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(x^2+8x+20)

\int \frac{x}{x ^{2} + 8 x + 20} d x

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

Solution steps

\int \frac{x}{x ^{2} + 8 x + 20} d x

Complete the square

x^{2} + 8 x + 20 : (x + 4)^{2} + 4

= \int \frac{x}{( x + 4 ) ^{2} + 4} d x

Apply u-substitution

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

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

Apply Integral 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 (\frac{x + 4}{2})^{2} + 1 - 2 arctan (\frac{x + 4}{2})

Simplify

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

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

Add a constant to the solution

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

\int e^{c o s (2 x)} sin (x) cos (x) d x

x \to + (- 1) lim (x^{2} + x + 1)

(t^{2} + x^{2}) d t + (2 t x - x) d x = 0

\frac{d}{d x} (x^{4} - 4 x^{2} - 10 x)

x \to 0 - lim (- 7 x^{2} + 5 x)

What is the integral of x/(x^2+8x+20) ?
The integral of x/(x^2+8x+20) is 1/2 ln| 1/4 (x+4)^2+1|-2arctan(1/2 (x+4))+C