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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 1 - 4 x

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

Simplify

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

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

Add a constant to the solution

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

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