What is the integral of (x^2-9x+5) ?

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

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integral of (x^2-9x+5)

\int (x^{2} - 9 x + 5) d x

\frac{x ^{3}}{3} - \frac{9 x ^{2}}{2} + 5 x + C

Solution steps

\int x^{2} - 9 x + 5 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^{2} d x - \int 9 x d x + \int 5 d x

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

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

\int 5 d x = 5 x

= \frac{x ^{3}}{3} - \frac{9 x ^{2}}{2} + 5 x

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{9 x ^{2}}{2} + 5 x + C

x \to + 1 lim (x^{2})

\frac{- 6 x ^{2} - 4 x + 4}{( 3 x ^{2} + 2 ) ^{2}}

\int \frac{sin ( y )}{cos ^{2} ( y )} d y

x \to 2 lim (\frac{1}{3})

x^{\frac{9}{6}}, 7

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