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

The integral from 1 to 2 of (x+1)/(x(x^2+1)) is -1/2 ln(5)+3/2 ln(2)-pi/4+arctan(2)

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

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

- \frac{1}{2} ln (5) + \frac{3}{2} ln (2) - \frac{π}{4} + arctan (2)

Solution steps

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

Take the partial fraction of

\frac{x + 1}{x ( x ^{2} + 1 )} : \frac{1}{x} + \frac{- x + 1}{x ^{2} + 1}

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

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

\int_{1}^{2} \frac{- x + 1}{x ^{2} + 1} d x = - \frac{1}{2} (ln (5) - ln (2)) + arctan (2) - \frac{π}{4}

= ln (2) - \frac{1}{2} (ln (5) - ln (2)) + arctan (2) - \frac{π}{4}

Simplify

ln (2) - \frac{1}{2} (ln (5) - ln (2)) + arctan (2) - \frac{π}{4} : - \frac{1}{2} ln (5) + \frac{3}{2} ln (2) - \frac{π}{4} + arctan (2)

= - \frac{1}{2} ln (5) + \frac{3}{2} ln (2) - \frac{π}{4} + arctan (2)

y = ln (x^{2} + 4)

x \to 6 lim (x)

\frac{\partial}{\partial x} (\frac{2}{x} - x)

\frac{d}{d x} (- 8 e^{- 2 x})

What is the integral from 1 to 2 of (x+1)/(x(x^2+1)) ?
The integral from 1 to 2 of (x+1)/(x(x^2+1)) is -1/2 ln(5)+3/2 ln(2)-pi/4+arctan(2)