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

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

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integral of (x-1)^2-4

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

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

Solution steps

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

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

\int 4 d x = 4 x

= \frac{x ^{3}}{3} - x^{2} + x - 4 x

Simplify

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

Add a constant to the solution

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

\int (x + 1) (x + 2) d x

x \to 1 lim (\frac{1}{x + 1})

f (x) = x^{2}, x = 0, x = 2

\frac{6 e ^{s} s - 5 e ^{s}}{( 6 s + 1 ) ^{2}}

\int \frac{x ^{3} + 22}{x ^{2} + 7 x + 12} d x

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