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

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

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

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

Solution steps

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

Expand

\frac{x + 4}{x ^{2} + 4 x + 6} : \frac{x}{x ^{2} + 4 x + 6} + \frac{4}{x ^{2} + 4 x + 6}

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

\int \frac{x}{x ^{2} + 4 x + 6} d x = \frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2))

\int \frac{4}{x ^{2} + 4 x + 6} d x = 22 arctan (\frac{1}{2} (x + 2))

= \frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2)) + 22 arctan (\frac{1}{2} (x + 2))

Simplify

\frac{1}{2} ln \frac{1}{2} (x + 2)^{2} + 1 - 2 arctan (\frac{1}{2} (x + 2)) + 22 arctan (\frac{1}{2} (x + 2)) : \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 + 2 arctan (\frac{1}{2} (x + 2))

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

Add a constant to the solution

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

\int e^{- 0.05 t} d t

\int x^{5} 13 + x^{6} d x

\int \frac{x ^{3} - x - 1}{x ^{2}} d x

\int x^{2} - 2 x + 2 d x

\int 4 x e^{- 3 x} d x

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