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

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

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integral of x(x^2+4)^{1/2}

\int x (x^{2} + 4)^{\frac{1}{2}} d x

\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int x (x^{2} + 4)^{\frac{1}{2}} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} + C

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