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

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

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

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

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

Solution steps

\int 3 x^{3} + (2 x - 3)^{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

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

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

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

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

Add a constant to the solution

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

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