What is the integral of (x^3)/((1-16x^2)^{3/2)} ?

The integral of (x^3)/((1-16x^2)^{3/2)} is (-8x^2+1)/(128(-16x^2+1)^{1/2)}+C

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integral of (x^3)/((1-16x^2)^{3/2)}

\int \frac{x ^{3}}{( 1 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{- 8 x ^{2} + 1}{128 ( - 16 x ^{2} + 1 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{3}}{( 1 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{u - 1}{512 u ^{\frac{3}{2}}} d u

= \int \frac{u - 1}{512 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}{512} \cdot \int \frac{u - 1}{u ^{\frac{3}{2}}} d u

Expand

\frac{u - 1}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{1}{u ^{\frac{3}{2}}}

= \frac{1}{512} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{512} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{1}{u ^{\frac{3}{2}}} d u)

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

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

= \frac{1}{512} (2 u^{\frac{1}{2}} - (- \frac{2}{u ^{\frac{1}{2}}}))

Substitute back

u = 1 - 16 x^{2}

= \frac{1}{512} (2 (1 - 16 x^{2})^{\frac{1}{2}} - (- \frac{2}{( 1 - 16 x ^{2} ) ^{\frac{1}{2}}}))

Simplify

\frac{1}{512} (2 (1 - 16 x^{2})^{\frac{1}{2}} - (- \frac{2}{( 1 - 16 x ^{2} ) ^{\frac{1}{2}}})) : \frac{- 8 x ^{2} + 1}{128 ( - 16 x ^{2} + 1 ) ^{\frac{1}{2}}}

= \frac{- 8 x ^{2} + 1}{128 ( - 16 x ^{2} + 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{- 8 x ^{2} + 1}{128 ( - 16 x ^{2} + 1 ) ^{\frac{1}{2}}} + C

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\frac{d}{d x} (x^{- 6 c o s (x)})

What is the integral of (x^3)/((1-16x^2)^{3/2)} ?
The integral of (x^3)/((1-16x^2)^{3/2)} is (-8x^2+1)/(128(-16x^2+1)^{1/2)}+C