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

The integral of x(1+4x^2)^{1/2} is 1/12 (1+4x^2)^{3/2}+C

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

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

\frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v ^{\frac{1}{2}}}{4} d v

= \frac{1}{2} \cdot \int \frac{v ^{\frac{1}{2}}}{4} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} (1 + 4 x^{2})^{\frac{3}{2}} : \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}}

= \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}} + C

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What is the integral of x(1+4x^2)^{1/2} ?
The integral of x(1+4x^2)^{1/2} is 1/12 (1+4x^2)^{3/2}+C