What is the integral of 18x^2sqrt(6x^3+5) ?

The integral of 18x^2sqrt(6x^3+5) is 2/3 (6x^3+5)^{3/2}+C

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integral of 18x^2sqrt(6x^3+5)

\int 18 x^{2} 6 x^{3} + 5 d x

\frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}} + C

Solution steps

\int 18 x^{2} 6 x^{3} + 5 d x

Take the constant out:

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

= 18 \cdot \int x^{2} 6 x^{3} + 5 d x

Apply u-substitution

: \int \frac{u ^{2}}{9} d u

= 18 \cdot \int \frac{u ^{2}}{9} d u

Take the constant out:

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

= 18 \cdot \frac{1}{9} \cdot \int u^{2} d u

Apply the Power Rule

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

= 18 \cdot \frac{1}{9} \cdot \frac{u ^{3}}{3}

Substitute back

u = 6 x^{3} + 5

= 18 \cdot \frac{1}{9} \cdot \frac{( 6 x ^{3} + 5 ) ^{3}}{3}

Simplify

18 \cdot \frac{1}{9} \cdot \frac{( 6 x ^{3} + 5 ) ^{3}}{3} : \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}}

= \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (6 x^{3} + 5)^{\frac{3}{2}} + C

\frac{\partial}{\partial y} (6 y^{2})

\int e^{- s t} \cdot sin (2 t) d t

x \to \frac{5 π}{2} lim (e^{t a n (x)})

y = 0, y = x, y = x - 2

\frac{\partial}{\partial x} ((3 x - 1)^{2} (x^{3} + 1)^{2})

What is the integral of 18x^2sqrt(6x^3+5) ?
The integral of 18x^2sqrt(6x^3+5) is 2/3 (6x^3+5)^{3/2}+C