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

The integral of (x^2)/(sqrt(3+1)) is (x^3)/6+C

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integral of (x^2)/(sqrt(3+1))

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

\frac{x ^{3}}{6} + C

Solution steps

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

Take the constant out:

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

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

\frac{1}{3 + 1} = \frac{1}{2}

= \frac{1}{2} \cdot \int x^{2} d x

Apply the Power Rule

= \frac{1}{2} \cdot \frac{x ^{3}}{3}

Simplify

\frac{1}{2} \cdot \frac{x ^{3}}{3} : \frac{x ^{3}}{6}

= \frac{x ^{3}}{6}

Add a constant to the solution

= \frac{x ^{3}}{6} + C

What is the integral of (x^2)/(sqrt(3+1)) ?
The integral of (x^2)/(sqrt(3+1)) is (x^3)/6+C