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

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

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

integral

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

\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 (x + 4)^{8}

\frac{- 1}{( 1 + x ) ^{2}}

ln^{5} (x^{2})

\frac{x ^{3}}{x ^{2}}

(e^{x})^{2} x

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