What is the integral of x/(16-x^2) ?

The integral of x/(16-x^2) is -1/2 ln|16-x^2|+C

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integral of x/(16-x^2)

\int \frac{x}{16 - x ^{2}} d x

- \frac{1}{2} ln 16 - x^{2} + C

Solution steps

\int \frac{x}{16 - 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 = 16 - x^{2}

= - \frac{1}{2} ln 16 - x^{2}

Add a constant to the solution

= - \frac{1}{2} ln 16 - x^{2} + C

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What is the integral of x/(16-x^2) ?
The integral of x/(16-x^2) is -1/2 ln|16-x^2|+C