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)

\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

\int \frac{sin ( x )}{cos ^{8} ( x )} d x

\int_{0}^{\frac{3 π}{2}} 4 ∣ sin (x) ∣ d x

f (x) = x^{2} e^{- 3 x}

\frac{d}{d x} (x + 7 x - 3)

3 y^{^{'}} + y = 0, y^{^{'}} (0) = 0

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