What is the integral of (3x+8)^{2/3} ?

The integral of (3x+8)^{2/3} is 1/5 (3x+8)^{5/3}+C

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integral of (3x+8)^{2/3}

\int (3 x + 8)^{\frac{2}{3}} d x

\frac{1}{5} (3 x + 8)^{\frac{5}{3}} + C

Solution steps

\int (3 x + 8)^{\frac{2}{3}} d x

Apply u-substitution

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

= \int \frac{u ^{\frac{2}{3}}}{3} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= \frac{1}{3} \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = 3 x + 8

= \frac{1}{3} \cdot \frac{3}{5} (3 x + 8)^{\frac{5}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{5} (3 x + 8)^{\frac{5}{3}} : \frac{1}{5} (3 x + 8)^{\frac{5}{3}}

= \frac{1}{5} (3 x + 8)^{\frac{5}{3}}

Add a constant to the solution

= \frac{1}{5} (3 x + 8)^{\frac{5}{3}} + C

\frac{d}{d x} (\frac{2}{( x - 1 ) ^{2}})

\int \frac{1 - sin ( x )}{1 + sin ( x )} d x

\frac{d r}{d θ} + r tan (θ) = 5 tan (θ)

y^{^{''''}} + y^{^{''}} - 6 y = 0

f (x) = sin (2 x)

What is the integral of (3x+8)^{2/3} ?
The integral of (3x+8)^{2/3} is 1/5 (3x+8)^{5/3}+C