What is the integral of (6x^2)(2x^3+1)^2 ?

The integral of (6x^2)(2x^3+1)^2 is 6((4x^9)/9+(2x^6)/3+(x^3)/3)+C

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integral of (6x^2)(2x^3+1)^2

\int (6 x^{2}) (2 x^{3} + 1)^{2} d x

6 (\frac{4 x ^{9}}{9} + \frac{2 x ^{6}}{3} + \frac{x ^{3}}{3}) + C

Solution steps

\int 6 x^{2} (2 x^{3} + 1)^{2} 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} + 1)^{2} d x

Expand

x^{2} (2 x^{3} + 1)^{2} : 4 x^{8} + 4 x^{5} + x^{2}

= 6 \cdot \int 4 x^{8} + 4 x^{5} + x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int 4 x^{8} d x + \int 4 x^{5} d x + \int x^{2} d x)

\int 4 x^{8} d x = \frac{4 x ^{9}}{9}

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int x^{2} d x = \frac{x ^{3}}{3}

= 6 (\frac{4 x ^{9}}{9} + \frac{2 x ^{6}}{3} + \frac{x ^{3}}{3})

Add a constant to the solution

= 6 (\frac{4 x ^{9}}{9} + \frac{2 x ^{6}}{3} + \frac{x ^{3}}{3}) + C

2 sec (θ)

\frac{d y}{d x} = - 5 y

\int_{0}^{9} 4 x + 5 - x d x

1 0^{t a n (π θ)}

f (x) = tan (2 x)

What is the integral of (6x^2)(2x^3+1)^2 ?
The integral of (6x^2)(2x^3+1)^2 is 6((4x^9)/9+(2x^6)/3+(x^3)/3)+C