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

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

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integral of x/(x^2-16)

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

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

Solution steps

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

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

Add a constant to the solution

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

- \frac{8}{x}

(x + y) \frac{d y}{d x} = 1

n = 1 \sum \infty n^{- 1}

- x^{2} + 9, - 2 x^{2} + 18

\int \frac{2 x ^{2} + 14 x + 8}{x ^{3} + 9 x} d x

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