What is the integral of 1/(x^2+3x+9) ?

The integral of 1/(x^2+3x+9) is 2/(3sqrt(3))arctan(2/(3sqrt(3))x+1/(sqrt(3)))+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+3x+9)

\int \frac{1}{x ^{2} + 3 x + 9} d x

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

Solution steps

\int \frac{1}{x ^{2} + 3 x + 9} d x

Complete the square

x^{2} + 3 x + 9 : (x + \frac{3}{2})^{2} + \frac{27}{4}

= \int \frac{1}{( x + \frac{3}{2} ) ^{2} + \frac{27}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 27} d u

= \int \frac{4}{4 u ^{2} + 27} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 27} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{6 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}{6 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}{6 3} arctan (v)

Substitute back

= 4 \cdot \frac{1}{6 3} arctan (\frac{2}{3 3} (x + \frac{3}{2}))

Simplify

4 \cdot \frac{1}{6 3} arctan (\frac{2}{3 3} (x + \frac{3}{2})) : \frac{2}{3 3} arctan (\frac{2}{3 3} x + \frac{1}{3})

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

Add a constant to the solution

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

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

\frac{\partial}{\partial y} (y^{3} + x y^{2} + y)

\int \frac{x ^{2}}{x ^{6} + 9} d x

\frac{\partial}{\partial x} (lo g_{e} (x^{2} + y^{2}))

y = x^{2}, y = 2 x

What is the integral of 1/(x^2+3x+9) ?
The integral of 1/(x^2+3x+9) is 2/(3sqrt(3))arctan(2/(3sqrt(3))x+1/(sqrt(3)))+C