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

The integral of sqrt((1+(1/(4x^{2/3))))} is 1/8 (4x^{2/3}+1)^{3/2}+C

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integral of sqrt((1+(1/(4x^{2/3))))}

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

\frac{1}{8} (4 x^{\frac{2}{3}} + 1)^{\frac{3}{2}} + C

Solution steps

\int 1 + \frac{1}{4 x ^{\frac{2}{3}}} d x

Apply u-substitution

: \int \frac{3 ( 4 u + 1 ) ^{\frac{1}{2}}}{4} d u

= \int \frac{3 ( 4 u + 1 ) ^{\frac{1}{2}}}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int (4 u + 1)^{\frac{1}{2}} d u

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{3}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} (4 x^{\frac{2}{3}} + 1)^{\frac{3}{2}} : \frac{1}{8} (4 x^{\frac{2}{3}} + 1)^{\frac{3}{2}}

= \frac{1}{8} (4 x^{\frac{2}{3}} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{8} (4 x^{\frac{2}{3}} + 1)^{\frac{3}{2}} + C

\frac{d y}{d x} = 2 y - 2 y^{5}

n = 0 \sum \infty \frac{1}{( n + 1 ) ^{2}}

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

f (x) = 5 e^{x} cos (x)

What is the integral of sqrt((1+(1/(4x^{2/3))))} ?
The integral of sqrt((1+(1/(4x^{2/3))))} is 1/8 (4x^{2/3}+1)^{3/2}+C