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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

= 2 (- \frac{1}{8} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{8} ln ∣ u ∣)

Substitute back

u = 9 - 4 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{12}{0.5 s ^{2} + 4 s + 5}

\int (x - 3) d x

\int_{0}^{\frac{1}{2}} 2 arcsin (2 y) d y

\int \frac{1}{x ( 49 + x )} d x

9 e^{(- 4 x)}

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