What is the integral of 4x^2(3x+1)^2 ?

The integral of 4x^2(3x+1)^2 is 4((9x^5)/5+(3x^4)/2+(x^3)/3)+C

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integral of 4x^2(3x+1)^2

\int 4 x^{2} (3 x + 1)^{2} d x

4 (\frac{9 x ^{5}}{5} + \frac{3 x ^{4}}{2} + \frac{x ^{3}}{3}) + C

Solution steps

\int 4 x^{2} (3 x + 1)^{2} d x

Take the constant out:

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

= 4 \cdot \int x^{2} (3 x + 1)^{2} d x

Expand

x^{2} (3 x + 1)^{2} : 9 x^{4} + 6 x^{3} + x^{2}

= 4 \cdot \int 9 x^{4} + 6 x^{3} + 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

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

\int 9 x^{4} d x = \frac{9 x ^{5}}{5}

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

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

= 4 (\frac{9 x ^{5}}{5} + \frac{3 x ^{4}}{2} + \frac{x ^{3}}{3})

Add a constant to the solution

= 4 (\frac{9 x ^{5}}{5} + \frac{3 x ^{4}}{2} + \frac{x ^{3}}{3}) + C

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What is the integral of 4x^2(3x+1)^2 ?
The integral of 4x^2(3x+1)^2 is 4((9x^5)/5+(3x^4)/2+(x^3)/3)+C