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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{5} - 2 x ^{2}}{( 1 + x ^{3} ) ^{\frac{3}{2}}} : \frac{x ^{5}}{( 1 + x ^{3} ) ^{\frac{3}{2}}} - \frac{2 x ^{2}}{( 1 + x ^{3} ) ^{\frac{3}{2}}}

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

Apply the Sum Rule:

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

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

\int \frac{x ^{5}}{( 1 + x ^{3} ) ^{\frac{3}{2}}} d x = \frac{2 x ^{3} + 4}{3 ( 1 + x ^{3} ) ^{\frac{1}{2}}}

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d u} (2 uv)

\frac{d}{d x} (ln ∣ coth (5 x) ∣)

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{3}})

n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n}

\int x^{3} e^{- 2 x} d x

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