What is the integral of 5x^2(2x^3+3)^4 ?

The integral of 5x^2(2x^3+3)^4 is 1/6 (2x^3+3)^5+C

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integral of 5x^2(2x^3+3)^4

\int 5 x^{2} (2 x^{3} + 3)^{4} d x

\frac{1}{6} (2 x^{3} + 3)^{5} + C

Solution steps

\int 5 x^{2} (2 x^{3} + 3)^{4} d x

Take the constant out:

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

= 5 \cdot \int x^{2} (2 x^{3} + 3)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{6} d u

= 5 \cdot \int \frac{u ^{4}}{6} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

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

= 5 \cdot \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x^{3} + 3

= 5 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 3 ) ^{5}}{5}

Simplify

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

= \frac{1}{6} (2 x^{3} + 3)^{5}

Add a constant to the solution

= \frac{1}{6} (2 x^{3} + 3)^{5} + C

\int \frac{( ln ( ln ( x ) ) )}{x} d x

\frac{\partial}{\partial y} (y e^{- x})

x \to - 2 lim (x^{2} + 5 x + 7)

\frac{1}{x ^{2}}, 5

What is the integral of 5x^2(2x^3+3)^4 ?
The integral of 5x^2(2x^3+3)^4 is 1/6 (2x^3+3)^5+C