What is the integral of [(x+3)(x-4)+1/(x^2)] ?

The integral of [(x+3)(x-4)+1/(x^2)] is (x^3)/3-(x^2)/2-12x-1/x+C

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integral of [(x+3)(x-4)+1/(x^2)]

\int [(x + 3) (x - 4) + \frac{1}{x ^{2}}] d x

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 12 x - \frac{1}{x} + C

Solution steps

\int (x + 3) (x - 4) + \frac{1}{x ^{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 + 3) (x - 4) d x + \int \frac{1}{x ^{2}} d x

\int (x + 3) (x - 4) d x = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 12 x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 12 x - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 12 x - \frac{1}{x} + C

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x \to 4 lim (\frac{1}{x - 3})

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What is the integral of [(x+3)(x-4)+1/(x^2)] ?
The integral of [(x+3)(x-4)+1/(x^2)] is (x^3)/3-(x^2)/2-12x-1/x+C