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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Expand

\frac{v - 4}{v ^{\frac{3}{2}}} : \frac{1}{v ^{\frac{1}{2}}} - \frac{4}{v ^{\frac{3}{2}}}

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

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

\int \frac{4}{v ^{\frac{3}{2}}} d v = - \frac{8}{v ^{\frac{1}{2}}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{6}{( s + 3 ) ^{2}}

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

x^{2}

\frac{2 s + 1}{s ^{2}}

x y^{^{'}} + y = x y^{3}

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