What is the integral of (x^3-1)/((2x^4-8x)^{3/2)} ?

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

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integral of (x^3-1)/((2x^4-8x)^{3/2)}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u ^{\frac{3}{2}}} d u

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

Apply the Power Rule

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

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

Substitute back

u = 2 x^{4} - 8 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (2 cos (x) + cos^{2} (x))

\frac{\partial}{\partial x} (\frac{c ^{2} - x ^{2}}{2 x})

\int \frac{1}{( x ^{2} + 25 ) ^{\frac{5}{2}}} d x

f (t) = e^{t a n (π t)}

y = \frac{6 x ^{2} + 2 x + 6}{x}

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