What is the integral of 1/x+x^2-2 ?

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

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integral of 1/x+x^2-2

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

ln ∣ x ∣ + \frac{x ^{3}}{3} - 2 x + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int 2 d x = 2 x

= ln ∣ x ∣ + \frac{x ^{3}}{3} - 2 x

Add a constant to the solution

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

\int_{- 1}^{0} \frac{4}{1 + ( 2 x + 1 ) ^{2}} d x

x^{3} + 9 x

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

f (x) = ln (1 - 3 x)

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