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

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

Popular >

integral of 1/((x^2+9))

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} arctan (u)

Substitute back

u = \frac{x}{3}

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} + x)

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{2}{x ^{4}}) (x + 8 x^{3}))

n = 1 \sum \infty (\frac{1}{2})^{\frac{1}{n}}

\int_{1}^{2} (x + \frac{1}{x})^{2} d x

\frac{\partial}{\partial y} (5 x^{2} - 3 x y + x yz)

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