What is the integral of (3x)/(1-4x^2) ?

The integral of (3x)/(1-4x^2) is -3/8 ln|1-4x^2|+C

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integral of (3x)/(1-4x^2)

\int \frac{3 x}{1 - 4 x ^{2}} d x

- \frac{3}{8} ln 1 - 4 x^{2} + C

Solution steps

\int \frac{3 x}{1 - 4 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{1 - 4 x ^{2}} d x

Apply u-substitution

= 3 \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

= 3 (- \frac{1}{8} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{8} ln ∣ u ∣)

Substitute back

u = 1 - 4 x^{2}

= 3 (- \frac{1}{8} ln 1 - 4 x^{2})

Simplify

3 (- \frac{1}{8} ln 1 - 4 x^{2}) : - \frac{3}{8} ln 1 - 4 x^{2}

= - \frac{3}{8} ln 1 - 4 x^{2}

Add a constant to the solution

= - \frac{3}{8} ln 1 - 4 x^{2} + C

What is the integral of (3x)/(1-4x^2) ?
The integral of (3x)/(1-4x^2) is -3/8 ln|1-4x^2|+C