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

The integral of (6x^2)(2x^3+5)^6 is 1/7 (2x^3+5)^7+C

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

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

\frac{1}{7} (2 x^{3} + 5)^{7} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{7}}{7}

= 6 \cdot \frac{1}{6} \cdot \frac{u ^{7}}{7}

Substitute back

u = 2 x^{3} + 5

= 6 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 5 ) ^{7}}{7}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 5 ) ^{7}}{7} : \frac{1}{7} (2 x^{3} + 5)^{7}

= \frac{1}{7} (2 x^{3} + 5)^{7}

Add a constant to the solution

= \frac{1}{7} (2 x^{3} + 5)^{7} + C

\int \frac{2 x + 7}{x ^{2} + 3 x + 3} d x

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{x ^{2} + 3}, y (1) = 9

(\frac{x ^{2}}{( x - 2 ) ^{2}})^{^{'}}

\int (sin (2 x)) d x

\frac{d y}{d x} = x^{2} - \frac{1}{2} y

What is the integral of (6x^2)(2x^3+5)^6 ?
The integral of (6x^2)(2x^3+5)^6 is 1/7 (2x^3+5)^7+C