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

The integral of 18x^2(2x^3+3)^5 is 1/2 (2x^3+3)^6+C

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

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

\frac{1}{2} (2 x^{3} + 3)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( 2 u ^{\frac{3}{2}} + 3 ) ^{5}}{2 x} d u

= 18 \cdot \int \frac{( 2 u + 3 ) ^{5}}{3} d u

Take the constant out:

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

= 18 \cdot \frac{1}{3} \cdot \int (2 u + 3)^{5} d u

Apply u-substitution

: \int \frac{v ^{5}}{2} d v

= 18 \cdot \frac{1}{3} \cdot \int \frac{v ^{5}}{2} d v

Take the constant out:

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

= 18 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= 18 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{v ^{6}}{6}

Substitute back

= 18 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{3} + 3 ) ^{6}}{6}

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} (2 x^{3} + 3)^{6} + C

x \to 7 lim (7 - 7)

4 \frac{d y}{d x} + \frac{7}{x y ^{3}} = 0, y (1) = 2

\frac{\partial}{\partial x} (\frac{3 x + 2 x y}{7 x + 6 y})

n = 1 \sum \infty (- \frac{2}{5})^{n}

\frac{1}{( s + 1 ) ( s ^{2} + 1 )}

What is the integral of 18x^2(2x^3+3)^5 ?
The integral of 18x^2(2x^3+3)^5 is 1/2 (2x^3+3)^6+C