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

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

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

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

\frac{1}{3} (x^{2} - 8) (x^{2} + 4)^{\frac{1}{2}} + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{3 ( u ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}} d u

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

Apply u-substitution

: \int \frac{3 ( v - 4 )}{2 v ^{\frac{1}{2}}} d v

= \frac{1}{3} \cdot \int \frac{3 ( v - 4 )}{2 v ^{\frac{1}{2}}} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int \frac{v - 4}{v ^{\frac{1}{2}}} d v

Expand

\frac{v - 4}{v ^{\frac{1}{2}}} : v^{\frac{1}{2}} - \frac{4}{v ^{\frac{1}{2}}}

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int v^{\frac{1}{2}} - \frac{4}{v ^{\frac{1}{2}}} d v

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} \cdot \frac{3}{2} (\int v^{\frac{1}{2}} d v - \int \frac{4}{v ^{\frac{1}{2}}} d v)

\int v^{\frac{1}{2}} d v = \frac{2}{3} v^{\frac{3}{2}}

\int \frac{4}{v ^{\frac{1}{2}}} d v = 8 v^{\frac{1}{2}}

= \frac{1}{3} \cdot \frac{3}{2} (\frac{2}{3} v^{\frac{3}{2}} - 8 v^{\frac{1}{2}})

Substitute back

= \frac{1}{3} \cdot \frac{3}{2} (\frac{2}{3} ((x^{3})^{\frac{2}{3}} + 4)^{\frac{3}{2}} - 8 ((x^{3})^{\frac{2}{3}} + 4)^{\frac{1}{2}})

Simplify

\frac{1}{3} \cdot \frac{3}{2} (\frac{2}{3} ((x^{3})^{\frac{2}{3}} + 4)^{\frac{3}{2}} - 8 ((x^{3})^{\frac{2}{3}} + 4)^{\frac{1}{2}}) : \frac{1}{3} (x^{2} - 8) (x^{2} + 4)^{\frac{1}{2}}

= \frac{1}{3} (x^{2} - 8) (x^{2} + 4)^{\frac{1}{2}}

Add a constant to the solution

= \frac{1}{3} (x^{2} - 8) (x^{2} + 4)^{\frac{1}{2}} + C

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