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

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

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

\int (x - 1) (3 x - 1) d x

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

Solution steps

\int (x - 1) (3 x - 1) d x

Expand

(x - 1) (3 x - 1) : 3 x^{2} - 4 x + 1

= \int 3 x^{2} - 4 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 3 x^{2} d x - \int 4 x d x + \int 1 d x

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

\int 4 x d x = 2 x^{2}

\int 1 d x = x

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

Add a constant to the solution

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

\int \frac{x e ^{x}}{e ^{x - 1}} d x

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

\int sin (t) d t

\int \frac{( 2 - x )}{x ^{2}} d x

\int (3 x + 1)^{6} d x

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