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

The integral of (5x^3+6)^4x^2 is 1/75 (5x^3+6)^5+C

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integral of (5x^3+6)^4x^2

\int (5 x^{3} + 6)^{4} x^{2} d x

\frac{1}{75} (5 x^{3} + 6)^{5} + C

Solution steps

\int (5 x^{3} + 6)^{4} x^{2} d x

Apply u-substitution

: \int \frac{( 5 u + 6 ) ^{4}}{3} d u

= \int \frac{( 5 u + 6 ) ^{4}}{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 (5 u + 6)^{4} d u

Apply u-substitution

: \int \frac{v ^{4}}{5} d v

= \frac{1}{3} \cdot \int \frac{v ^{4}}{5} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{5} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

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

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{5} \cdot \frac{( 5 x ^{3} + 6 ) ^{5}}{5} : \frac{1}{75} (5 x^{3} + 6)^{5}

= \frac{1}{75} (5 x^{3} + 6)^{5}

Add a constant to the solution

= \frac{1}{75} (5 x^{3} + 6)^{5} + C

y^{^{'}} - y = 5 t e^{2 t}, y (0) = 1

n = 0 \sum \infty (1.075)^{n}

y = x + 1, (0, 2)

7 x^{3} - 6 x + \frac{3}{x ^{7}} + 9

x \to + (- 4) lim (x^{2} - 3)

What is the integral of (5x^3+6)^4x^2 ?
The integral of (5x^3+6)^4x^2 is 1/75 (5x^3+6)^5+C