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

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

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integral of ((x^2+1))/(x^2)

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

- \frac{1}{x} + x + C

Solution steps

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

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

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

\int 1 d x = x

= - \frac{1}{x} + x

Add a constant to the solution

= - \frac{1}{x} + x + C

\frac{d}{d x} (arccsc (e^{x}))

\frac{\partial}{\partial y} (y^{4} cos (2 x))

y = x^{0.85}

\frac{d v}{d t} = - b v

(x + 2)^{3}

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