What is the integral of sqrt(1+4/(9x^{2/3))} ?

The integral of sqrt(1+4/(9x^{2/3))} is 1/27 (9x^{2/3}+4)^{3/2}+C

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integral of sqrt(1+4/(9x^{2/3))}

\int 1 + \frac{4}{9 x ^{\frac{2}{3}}} d x

\frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} + C

Solution steps

\int 1 + \frac{4}{9 x ^{\frac{2}{3}}} d x

Simplify

1 + \frac{4}{9 x ^{\frac{2}{3}}} : \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}}

= \int \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\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}{3} \cdot \int \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{6} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 9 x^{\frac{2}{3}} + 4

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} : \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} + C

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What is the integral of sqrt(1+4/(9x^{2/3))} ?
The integral of sqrt(1+4/(9x^{2/3))} is 1/27 (9x^{2/3}+4)^{3/2}+C