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

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

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

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

\frac{1}{3} x^{3} + x^{6} + x^{9} + C

Solution steps

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

Apply u-substitution

= \int \frac{( 1 + 3 u ) ^{2}}{3} d u

Take the constant out:

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

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

Expand

(1 + 3 u)^{2} : 1 + 6 u + 9 u^{2}

= \frac{1}{3} \cdot \int 1 + 6 u + 9 u^{2} d u

Apply the Sum Rule:

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

= \frac{1}{3} (\int 1 d u + \int 6 u d u + \int 9 u^{2} d u)

\int 1 d u = u

\int 6 u d u = 3 u^{2}

\int 9 u^{2} d u = 3 u^{3}

= \frac{1}{3} (u + 3 u^{2} + 3 u^{3})

Substitute back

u = x^{3}

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

Simplify

\frac{1}{3} (x^{3} + 3 (x^{3})^{2} + 3 (x^{3})^{3}) : \frac{1}{3} x^{3} + x^{6} + x^{9}

= \frac{1}{3} x^{3} + x^{6} + x^{9}

Add a constant to the solution

= \frac{1}{3} x^{3} + x^{6} + x^{9} + C

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