What is the integral of 5/((x^2+6x+11)) ?

The integral of 5/((x^2+6x+11)) is 5/(sqrt(2))arctan(1/(sqrt(2))(x+3))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/((x^2+6x+11))

\int \frac{5}{( x ^{2} + 6 x + 11 )} d x

\frac{5}{2} arctan (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{5}{x ^{2} + 6 x + 11} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} + 6 x + 11} d x

Complete the square

x^{2} + 6 x + 11 : (x + 3)^{2} + 2

= 5 \cdot \int \frac{1}{( x + 3 ) ^{2} + 2} d x

Apply u-substitution

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

= 5 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= 5 \cdot \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

= 5 \cdot \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)

= 5 \cdot \frac{1}{2} arctan (v)

Substitute back

= 5 \cdot \frac{1}{2} arctan (\frac{x + 3}{2})

Simplify

5 \cdot \frac{1}{2} arctan (\frac{x + 3}{2}) : \frac{5}{2} arctan (\frac{1}{2} (x + 3))

= \frac{5}{2} arctan (\frac{1}{2} (x + 3))

Add a constant to the solution

= \frac{5}{2} arctan (\frac{1}{2} (x + 3)) + C

\int x + a d x

\int \frac{1}{2} t^{- 3} (5 t^{- 2} + 10)^{- 4} d t

\int \frac{cos ( 19 x )}{sin ^{2} ( 19 x )} d x

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

\int (x^{2} + 1) ln (2 x) d x

What is the integral of 5/((x^2+6x+11)) ?
The integral of 5/((x^2+6x+11)) is 5/(sqrt(2))arctan(1/(sqrt(2))(x+3))+C