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

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

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

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

ln ∣ x ∣ - \frac{y ^{2}}{x} + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{y ^{2}}{x}

Add a constant to the solution

= ln ∣ x ∣ - \frac{y ^{2}}{x} + C

\frac{d}{d x} (ln ((x - 1)^{3} (x^{2} + 1)^{4}))

\frac{d}{d x} (3 x^{2} - e^{2 x})

\frac{d y}{d t} = - \frac{3 y}{t} - 2 - t^{- 4}

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

\int sin^{3} (x) cos^{2} (x) d x

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