What is the integral of 2x^3-6x^2+2x-1+(2/x) ?

The integral of 2x^3-6x^2+2x-1+(2/x) is (x^4)/2-2x^3+x^2-x+2ln|x|+C

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integral of 2x^3-6x^2+2x-1+(2/x)

\int 2 x^{3} - 6 x^{2} + 2 x - 1 + (\frac{2}{x}) d x

\frac{x ^{4}}{2} - 2 x^{3} + x^{2} - x + 2 ln ∣ x ∣ + C

Solution steps

\int 2 x^{3} - 6 x^{2} + 2 x - 1 + \frac{2}{x} 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^{3} d x - \int 6 x^{2} d x + \int 2 x d x - \int 1 d x + \int \frac{2}{x} d x

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 6 x^{2} d x = 2 x^{3}

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

\int 1 d x = x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (sin (x y) + cos (x y))

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

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

y^{^{'}} + 5 y = 4, y (0) = 0

\frac{d}{d x} (\frac{8 x ^{2} + 302 x}{x - 10})

What is the integral of 2x^3-6x^2+2x-1+(2/x) ?
The integral of 2x^3-6x^2+2x-1+(2/x) is (x^4)/2-2x^3+x^2-x+2ln|x|+C