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

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

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

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

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

Solution steps

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

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

\int 24 sec^{4} (x) d x

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f (x) = \frac{2 x}{e ^{3 x^{2}}}, at x = - 1

\int \frac{1}{2 + 3 y ^{2}} d y

\int \frac{6}{2 + 3 x} d x

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