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

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

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

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

\frac{x ^{9}}{3} + x^{6} + x^{3} + C

Solution steps

\int 3 x^{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

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

Expand

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

= 3 \cdot \int x^{8} + 2 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

= 3 (\int x^{8} d x + \int 2 x^{5} d x + \int x^{2} d x)

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

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

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

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

Simplify

3 (\frac{x ^{9}}{9} + \frac{x ^{6}}{3} + \frac{x ^{3}}{3}) : \frac{x ^{9}}{3} + x^{6} + x^{3}

= \frac{x ^{9}}{3} + x^{6} + x^{3}

Add a constant to the solution

= \frac{x ^{9}}{3} + x^{6} + x^{3} + C

\frac{\partial}{\partial x} (cos (y) + y e^{x})

\int \frac{16}{x ^{2} 16 - x ^{2}} d x

y^{^{'}} = - \frac{y}{( x + \frac{2}{y} )}

y^{^{'}} + y = b

x + 2

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