What is the integral of (4x^2+9x) ?

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

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integral of (4x^2+9x)

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

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

Solution steps

\int 4 x^{2} + 9 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 4 x^{2} d x + \int 9 x d x

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

\int 9 x d x = \frac{9 x ^{2}}{2}

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

Add a constant to the solution

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

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