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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

Expand

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

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

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

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

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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