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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4 u ^{3}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 1 - 4 x

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

Simplify

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

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

Add a constant to the solution

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

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