What is the integral of x/(8x^2+6) ?

The integral of x/(8x^2+6) is 1/16 ln|8x^2+6|+C

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integral of x/(8x^2+6)

\int \frac{x}{8 x ^{2} + 6} d x

\frac{1}{16} ln 8 x^{2} + 6 + C

Solution steps

\int \frac{x}{8 x ^{2} + 6} d x

Apply u-substitution

= \int \frac{1}{16 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{16} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{16} ln ∣ u ∣

Substitute back

u = 8 x^{2} + 6

= \frac{1}{16} ln 8 x^{2} + 6

Add a constant to the solution

= \frac{1}{16} ln 8 x^{2} + 6 + C

What is the integral of x/(8x^2+6) ?
The integral of x/(8x^2+6) is 1/16 ln|8x^2+6|+C