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

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

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integral of x/(9-x^2)

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

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

Solution steps

\int \frac{x}{9 - x ^{2}} 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 = 9 - x^{2}

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

Add a constant to the solution

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

\int \frac{1}{sin ^{2} ( 3 x )} d x

\frac{d y}{d x} = \frac{sec ( y )}{( x - 3 ) ^{2}}

\frac{d y}{d x} - 2 x y = x y^{2}

\frac{7 - sec ( x )}{tan ( x )}

\int_{0}^{1} 8 cos (x^{2}) d x

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