What is the integral of x/(x^2-8) ?

The integral of x/(x^2-8) is 1/2 ln|x^2-8|+C

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integral of x/(x^2-8)

\int \frac{x}{x ^{2} - 8} d x

\frac{1}{2} ln x^{2} - 8 + C

Solution steps

\int \frac{x}{x ^{2} - 8} 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} - 8

= \frac{1}{2} ln x^{2} - 8

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 8 + C

q^{^{''}} + 2 q^{^{'}} + 4 q = 50 cos (t)

1 + x

y^{^{'}} = 4 x y^{3}

cos^{2} (x) sin (x) \frac{d y}{d x} + (cos^{3} (x)) y = 1

x \to - \infty lim (x^{2} - 2 x + 1 - x)

What is the integral of x/(x^2-8) ?
The integral of x/(x^2-8) is 1/2 ln|x^2-8|+C