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

= 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

= 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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (e^{x})

\int_{1}^{2} \frac{3}{4 x - x ^{2}} d x

t an g e n t f (x) = x - 4, at x = 13

y^{^{'}} = - \frac{2 x}{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