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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{8 u ^{2}} d u

= 3 \cdot \int - \frac{1}{8 u ^{2}} 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 ^{2}} d u)

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 1 - 4 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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