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

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

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

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

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

Solution steps

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

Expand

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

= 3 \cdot \int x^{8} + 4 x^{5} + 4 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 4 x^{5} d x + \int 4 x^{2} d x)

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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