What is the integral of x^3sqrt(a^2+2x^2) ?

The integral of x^3sqrt(a^2+2x^2) is 1/30 (3x^2-a^2)(2x^2+a^2)^{3/2}+C

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integral of x^3sqrt(a^2+2x^2)

\int x^{3} a^{2} + 2 x^{2} d x

\frac{1}{30} (3 x^{2} - a^{2}) (2 x^{2} + a^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} a^{2} + 2 x^{2} d x

Apply u-substitution

: \int \frac{u ( u - a ^{2} )}{8} d u

= \int \frac{u ( u - a ^{2} )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u (u - a^{2}) d u

Expand

u (u - a^{2}) : u^{\frac{3}{2}} - a^{2} u

= \frac{1}{8} \cdot \int u^{\frac{3}{2}} - a^{2} u 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}{8} (\int u^{\frac{3}{2}} d u - \int a^{2} u d u)

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

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

= \frac{1}{8} (\frac{2}{5} u^{\frac{5}{2}} - a^{2} \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = a^{2} + 2 x^{2}

= \frac{1}{8} (\frac{2}{5} (a^{2} + 2 x^{2})^{\frac{5}{2}} - a^{2} \frac{2}{3} (a^{2} + 2 x^{2})^{\frac{3}{2}})

Simplify

\frac{1}{8} (\frac{2}{5} (a^{2} + 2 x^{2})^{\frac{5}{2}} - a^{2} \frac{2}{3} (a^{2} + 2 x^{2})^{\frac{3}{2}}) : \frac{1}{30} (3 x^{2} - a^{2}) (2 x^{2} + a^{2})^{\frac{3}{2}}

= \frac{1}{30} (3 x^{2} - a^{2}) (2 x^{2} + a^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{30} (3 x^{2} - a^{2}) (2 x^{2} + a^{2})^{\frac{3}{2}} + C

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What is the integral of x^3sqrt(a^2+2x^2) ?
The integral of x^3sqrt(a^2+2x^2) is 1/30 (3x^2-a^2)(2x^2+a^2)^{3/2}+C