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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{4}{3} arctan (\frac{1}{3} (x + 3)) + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 6 x + 18} d x

Complete the square

x^{2} + 6 x + 18 : (x + 3)^{2} + 9

= 4 \cdot \int \frac{1}{( x + 3 ) ^{2} + 9} d x

Apply u-substitution

= 4 \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{3} arctan (v)

Substitute back

= 4 \cdot \frac{1}{3} arctan (\frac{x + 3}{3})

Simplify

4 \cdot \frac{1}{3} arctan (\frac{x + 3}{3}) : \frac{4}{3} arctan (\frac{1}{3} (x + 3))

= \frac{4}{3} arctan (\frac{1}{3} (x + 3))

Add a constant to the solution

= \frac{4}{3} arctan (\frac{1}{3} (x + 3)) + C

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