What is the integral of x/(4x^2+9) ?

The integral of x/(4x^2+9) is 1/8 ln|4x^2+9|+C

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integral of x/(4x^2+9)

\int \frac{x}{4 x ^{2} + 9} d x

\frac{1}{8} ln 4 x^{2} + 9 + C

Solution steps

\int \frac{x}{4 x ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 x^{2} + 9

= \frac{1}{8} ln 4 x^{2} + 9

Add a constant to the solution

= \frac{1}{8} ln 4 x^{2} + 9 + C

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What is the integral of x/(4x^2+9) ?
The integral of x/(4x^2+9) is 1/8 ln|4x^2+9|+C