What is the integral of 1/(x^2+4x+8) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2+4x+8)

\int \frac{1}{x ^{2} + 4 x + 8} d x

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

Solution steps

\int \frac{1}{x ^{2} + 4 x + 8} d x

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= \frac{1}{2} arctan (v)

Substitute back

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

Add a constant to the solution

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

\frac{d}{d x} (50 - x^{2})

\frac{d}{d x} ((\frac{3}{4} x)^{4})

x \to 0 lim ((3 x + 1)^{\frac{1}{x}})

\frac{\partial}{\partial y} (x y sin (x))

\frac{\partial}{\partial x} (y^{4} - 4 x y)

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