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

The integral of x(x-3)^2+4^2 is (x^4)/4-2x^3+(9x^2)/2+16x+C

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

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

\frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + 16 x + C

Solution steps

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

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

\int 4^{2} d x = 16 x

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + 16 x

Add a constant to the solution

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + 16 x + C

y = x^{2}, y = x + 2, x = - 2

ln (\frac{1}{x + 5}) + x cos (x)

\int sin (6 π t) d t

y = x - x^{3}, (1, 0)

\int 1 + tan (x) d x

What is the integral of x(x-3)^2+4^2 ?
The integral of x(x-3)^2+4^2 is (x^4)/4-2x^3+(9x^2)/2+16x+C