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

The integral of x/((x^2+9)) is 1/2 ln|x^2+9|+C

Popular >

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

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

\frac{1}{2} ln x^{2} + 9 + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 u} d u

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 9

= \frac{1}{2} ln x^{2} + 9

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 9 + C

x - 5

\int \frac{g}{\frac{1}{p} - \frac{g z}{k}} d z

\int_{0}^{2} 0.2 e^{- 0.2 x} d x

y^{^{'}} - 5 y = t e^{(- 4 t)}

f (x) = \frac{1}{x ^{2}}, at x = 2

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