What is the integral of (8x)/(4x^2-9) ?

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

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integral of (8x)/(4x^2-9)

\int \frac{8 x}{4 x ^{2} - 9} d x

ln 4 x^{2} - 9 + C

Solution steps

\int \frac{8 x}{4 x ^{2} - 9} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{4 x ^{2} - 9} d x

Apply u-substitution

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

= 8 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 x^{2} - 9

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

Simplify

8 \cdot \frac{1}{8} ln 4 x^{2} - 9 : ln 4 x^{2} - 9

= ln 4 x^{2} - 9

Add a constant to the solution

= ln 4 x^{2} - 9 + C

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\int 15 x e^{- 2 x} d x

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