What is the integral from 1 to 2 of (x/y-y/x) ?

The integral from 1 to 2 of (x/y-y/x) is xln(2)-3/(2x)

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integral from 1 to 2 of (x/y-y/x)

\int_{1}^{2} (\frac{x}{y} - \frac{y}{x}) d y

x ln (2) - \frac{3}{2 x}

Solution steps

\int_{1}^{2} \frac{x}{y} - \frac{y}{x} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{x}{y} d y - \int_{1}^{2} \frac{y}{x} d y

\int_{1}^{2} \frac{x}{y} d y = x ln (2)

\int_{1}^{2} \frac{y}{x} d y = \frac{3}{2 x}

= x ln (2) - \frac{3}{2 x}

\int_{0}^{1} h (x - x^{2}) d x

\int_{2 x}^{x^{2}} t^{2} + 1 d t

\int_{0}^{\frac{1}{3}} x ln (3 x) d x

\int_{2}^{5} \frac{1}{( x - 2 ) ^{4}} d x

\int_{0}^{\frac{π}{2}} x cos (7 x) d x

What is the integral from 1 to 2 of (x/y-y/x) ?
The integral from 1 to 2 of (x/y-y/x) is xln(2)-3/(2x)