What is the integral of-1/(Ux)+1/(U(x-U)) ?

The integral of-1/(Ux)+1/(U(x-U)) is 1/U ln|-U+x|-1/U ln|x|+C

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integral of-1/(Ux)+1/(U(x-U))

\int - \frac{1}{Ux} + \frac{1}{U ( x - U )} d x

\frac{1}{U} ln ∣ - U + x ∣ - \frac{1}{U} ln ∣ x ∣ + C

Solution steps

\int - \frac{1}{Ux} + \frac{1}{U ( x - U )} 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}{( - U + x ) U} d x - \int \frac{1}{Ux} d x

\int \frac{1}{( - U + x ) U} d x = \frac{1}{U} ln ∣ - U + x ∣

\int \frac{1}{Ux} d x = \frac{1}{U} ln ∣ x ∣

= \frac{1}{U} ln ∣ - U + x ∣ - \frac{1}{U} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{U} ln ∣ - U + x ∣ - \frac{1}{U} ln ∣ x ∣ + C

\frac{\partial}{\partial y} (\frac{e ^{y}}{x + y ^{2}})

\frac{d y}{d x} = 17

\frac{\partial}{\partial x} (y cos (π x))

\int t^{6} ln (t) d t

What is the integral of-1/(Ux)+1/(U(x-U)) ?
The integral of-1/(Ux)+1/(U(x-U)) is 1/U ln|-U+x|-1/U ln|x|+C