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

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

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integral of x(3x+1)^2

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

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

Solution steps

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

Expand

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

= \int 9 x^{3} + 6 x^{2} + x 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

= \int 9 x^{3} d x + \int 6 x^{2} d x + \int x d x

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

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

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

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

Add a constant to the solution

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

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