What is the integral of x/(x^{2_{-1)}} ?

The integral of x/(x^{2_{-1)}} is (x^{-2_{-1}+2})/(-2_{-1)+2}+C

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integral of x/(x^{2_{-1)}}

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

\frac{x ^{- 2_{- 1} + 2}}{- 2 _{- 1} + 2} + C

Solution steps

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

Simplify

\frac{x}{x ^{2_{- 1}}} : x^{1 - 2_{- 1}}

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

Apply the Power Rule

= \frac{x ^{- 2_{- 1} + 2}}{- 2 _{- 1} + 2}

Add a constant to the solution

= \frac{x ^{- 2_{- 1} + 2}}{- 2 _{- 1} + 2} + C

\int_{0}^{1} (3 x^{2} + 2 x + 1) d x

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

\frac{\partial}{\partial x} (tan (x + y))

\int ln (r^{2}) d r

y^{^{'}} = 5 y sin (4 x)

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