What is the integral of (x^2)/2 (2+(x^3)/6)^{1/2} ?

The integral of (x^2)/2 (2+(x^3)/6)^{1/2} is 1/(9*6^{1/2)}(12+x^3)^{3/2}+C

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integral of (x^2)/2 (2+(x^3)/6)^{1/2}

\int \frac{x ^{2}}{2} (2 + \frac{x ^{3}}{6})^{\frac{1}{2}} d x

\frac{1}{9 \cdot 6 ^{\frac{1}{2}}} (12 + x^{3})^{\frac{3}{2}} + C

Solution steps

\int \frac{x ^{2}}{2} (2 + \frac{x ^{3}}{6})^{\frac{1}{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{2} (2 + \frac{x ^{3}}{6})^{\frac{1}{2}} d x

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{( 12 + u ) ^{\frac{1}{2}}}{3 \cdot 6 ^{\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 \frac{1}{3 \cdot 6 ^{\frac{1}{2}}} \cdot \int (12 + u)^{\frac{1}{2}} d u

Apply u-substitution

= \frac{1}{2} \cdot \frac{1}{3 \cdot 6 ^{\frac{1}{2}}} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

= \frac{1}{2} \cdot \frac{1}{3 \cdot 6 ^{\frac{1}{2}}} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= \frac{1}{2} \cdot \frac{1}{3 \cdot 6 ^{\frac{1}{2}}} \cdot \frac{2}{3} (12 + x^{3})^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{3 \cdot 6 ^{\frac{1}{2}}} \cdot \frac{2}{3} (12 + x^{3})^{\frac{3}{2}} : \frac{1}{9 \cdot 6 ^{\frac{1}{2}}} (12 + x^{3})^{\frac{3}{2}}

= \frac{1}{9 \cdot 6 ^{\frac{1}{2}}} (12 + x^{3})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9 \cdot 6 ^{\frac{1}{2}}} (12 + x^{3})^{\frac{3}{2}} + C

What is the integral of (x^2)/2 (2+(x^3)/6)^{1/2} ?
The integral of (x^2)/2 (2+(x^3)/6)^{1/2} is 1/(9*6^{1/2)}(12+x^3)^{3/2}+C