What is the integral of (x^2)/4-1 ?

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

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integral of (x^2)/4-1

\int \frac{x ^{2}}{4} - 1 d x

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

Solution steps

\int \frac{x ^{2}}{4} - 1 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 \frac{x ^{2}}{4} d x - \int 1 d x

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

\int 1 d x = x

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

Add a constant to the solution

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

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