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

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

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integral of (x+1)^2-1

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

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

Solution steps

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

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

\int 1 d x = x

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

Simplify

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

Add a constant to the solution

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