What is the integral of x(4x+1)^2 ?

The integral of x(4x+1)^2 is 4x^4+(8x^3)/3+(x^2)/2+C

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integral of x(4x+1)^2

\int x (4 x + 1)^{2} d x

4 x^{4} + \frac{8 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Solution steps

\int x (4 x + 1)^{2} d x

Expand

x (4 x + 1)^{2} : 16 x^{3} + 8 x^{2} + x

= \int 16 x^{3} + 8 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 16 x^{3} d x + \int 8 x^{2} d x + \int x d x

\int 16 x^{3} d x = 4 x^{4}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

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

= 4 x^{4} + \frac{8 x ^{3}}{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= 4 x^{4} + \frac{8 x ^{3}}{3} + \frac{x ^{2}}{2} + C

What is the integral of x(4x+1)^2 ?
The integral of x(4x+1)^2 is 4x^4+(8x^3)/3+(x^2)/2+C