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

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

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integral of (x+1)(2x-1)

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

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

Solution steps

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

Expand

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{1}{x + 6}

\int \frac{1}{x ( ln ( x ) ) ^{15}} d x

\frac{d}{d x} ((2 x^{3} - 7)^{8})

\frac{d}{d x} (ln (\frac{x}{2}) (x - 2))

x \to 0 - lim (\frac{x + 10}{x ^{2}})

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