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

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

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

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

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + arctan (x) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{- x + 1}{x ^{2} + 1} d x = - \frac{1}{2} ln x^{2} + 1 + arctan (x)

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + arctan (x) + C

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