What is the integral of (x^5)/((1+x^3)^{5/2)} ?

The integral of (x^5)/((1+x^3)^{5/2)} is (-6x^3-4)/(9(x^3+1)^{3/2)}+C

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integral of (x^5)/((1+x^3)^{5/2)}

\int \frac{x ^{5}}{( 1 + x ^{3} ) ^{\frac{5}{2}}} d x

\frac{- 6 x ^{3} - 4}{9 ( x ^{3} + 1 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{x ^{5}}{( 1 + x ^{3} ) ^{\frac{5}{2}}} d x

Apply u-substitution

: \int \frac{u - 1}{3 u ^{\frac{5}{2}}} d u

= \int \frac{u - 1}{3 u ^{\frac{5}{2}}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{u - 1}{u ^{\frac{5}{2}}} d u

Expand

\frac{u - 1}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{3}{2}}} - \frac{1}{u ^{\frac{5}{2}}}

= \frac{1}{3} \cdot \int \frac{1}{u ^{\frac{3}{2}}} - \frac{1}{u ^{\frac{5}{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}{3} (\int \frac{1}{u ^{\frac{3}{2}}} d u - \int \frac{1}{u ^{\frac{5}{2}}} d u)

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

\int \frac{1}{u ^{\frac{5}{2}}} d u = - \frac{2}{3 u ^{\frac{3}{2}}}

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

Substitute back

u = 1 + x^{3}

= \frac{1}{3} (- \frac{2}{( 1 + x ^{3} ) ^{\frac{1}{2}}} - (- \frac{2}{3 ( 1 + x ^{3} ) ^{\frac{3}{2}}}))

Simplify

\frac{1}{3} (- \frac{2}{( 1 + x ^{3} ) ^{\frac{1}{2}}} - (- \frac{2}{3 ( 1 + x ^{3} ) ^{\frac{3}{2}}})) : \frac{- 6 x ^{3} - 4}{9 ( x ^{3} + 1 ) ^{\frac{3}{2}}}

= \frac{- 6 x ^{3} - 4}{9 ( x ^{3} + 1 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{- 6 x ^{3} - 4}{9 ( x ^{3} + 1 ) ^{\frac{3}{2}}} + C

\int cos (x - n x) d x

\int e^{- x} x^{3} d x

y^{^{''}} + 4 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 13

9 (5^{x})

y^{^{'}} = 9 x y

What is the integral of (x^5)/((1+x^3)^{5/2)} ?
The integral of (x^5)/((1+x^3)^{5/2)} is (-6x^3-4)/(9(x^3+1)^{3/2)}+C