What is the integral of x^3(1+9x^4)^{-3/2} ?

The integral of x^3(1+9x^4)^{-3/2} is -1/(18(9x^4+1)^{1/2)}+C

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integral of x^3(1+9x^4)^{-3/2}

\int x^{3} (1 + 9 x^{4})^{- \frac{3}{2}} d x

- \frac{1}{18 ( 9 x ^{4} + 1 ) ^{\frac{1}{2}}} + C

Solution steps

\int x^{3} (1 + 9 x^{4})^{- \frac{3}{2}} d x

Apply u-substitution

: \int \frac{1}{36 u ^{\frac{3}{2}}} d u

= \int \frac{1}{36 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}{36} \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

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

= \frac{1}{36} (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = 1 + 9 x^{4}

= \frac{1}{36} (- \frac{2}{( 1 + 9 x ^{4} ) ^{\frac{1}{2}}})

Simplify

\frac{1}{36} (- \frac{2}{( 1 + 9 x ^{4} ) ^{\frac{1}{2}}}) : - \frac{1}{18 ( 9 x ^{4} + 1 ) ^{\frac{1}{2}}}

= - \frac{1}{18 ( 9 x ^{4} + 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{1}{18 ( 9 x ^{4} + 1 ) ^{\frac{1}{2}}} + C

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What is the integral of x^3(1+9x^4)^{-3/2} ?
The integral of x^3(1+9x^4)^{-3/2} is -1/(18(9x^4+1)^{1/2)}+C