What is the integral of y=x^2+x-12 ?

The integral of y=x^2+x-12 is (x^3)/3+(x^2)/2-12x+C

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integral of y=x^2+x-12

integral

y = x^{2} + x - 12

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 12 x + C

Solution steps

\int x^{2} + x - 12 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 x d x - \int 12 d x

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

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

\int 12 d x = 12 x

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 12 x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 12 x + C

in t e g r a l \frac{x + 1}{( x ^{2} + 4 x + 8 )}

in t e g r a l \frac{u}{( 1 + u )}

in t e g r a l 7^{2} x + 3

in t e g r a l 4 x - 1

in t e g r a l \frac{( 1 + x )}{x ^{2}}

What is the integral of y=x^2+x-12 ?
The integral of y=x^2+x-12 is (x^3)/3+(x^2)/2-12x+C