What is the integral of (x-1)^3-(x-1) ?

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

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integral of (x-1)^3-(x-1)

\int (x - 1)^{3} - (x - 1) d x

\frac{( x - 1 ) ^{4}}{4} - \frac{x ^{2}}{2} + x + C

Solution steps

\int (x - 1)^{3} - (x - 1) d x

Simplify

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

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

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

\int 1 d x = x

= \frac{( x - 1 ) ^{4}}{4} - \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{( x - 1 ) ^{4}}{4} - \frac{x ^{2}}{2} + x + C

\int \frac{3}{2 x ^{2} + 8 x + 10} d x

\int (e^{x} + 1) e^{x} d x

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

\int \frac{x}{x ^{4} - 2 x ^{2} + 10} d x

\int 2 x^{3} x^{2} + 2 d x

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