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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4 u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

= 3 (- \frac{1}{4} (- \frac{1}{u}))

Substitute back

u = 1 - 4 x

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

Simplify

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

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

Add a constant to the solution

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

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