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

The integral of 1/(x(x^2+α)) is 1/(2α)(-ln((|x^2+α|}{|α|})+ln|\frac{x^2+α)/α-1|)+C

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integral of 1/(x(x^2+α))

\int \frac{1}{x ( x ^{2} + α )} d x

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

Solution steps

\int \frac{1}{x ( x ^{2} + α )} d x

Apply u-substitution

= \int \frac{1}{2 u ( u + α )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ( u + α )} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{1}{( v - α ) v} d v

v - α = α (\frac{v}{α} - 1)

= \frac{1}{2} \cdot \int \frac{1}{α ( \frac{v}{α} - 1 ) v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{α} \cdot \int \frac{1}{( \frac{v}{α} - 1 ) v} d v

Apply u-substitution

= \frac{1}{2} \cdot \frac{1}{α} \cdot \int \frac{1}{w ( w - 1 )} d w

Take the partial fraction of

\frac{1}{w ( w - 1 )} : - \frac{1}{w} + \frac{1}{w - 1}

= \frac{1}{2} \cdot \frac{1}{α} \cdot \int - \frac{1}{w} + \frac{1}{w - 1} d w

Apply the Sum Rule:

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

= \frac{1}{2} \cdot \frac{1}{α} (- \int \frac{1}{w} d w + \int \frac{1}{w - 1} d w)

\int \frac{1}{w} d w = ln ∣ w ∣

\int \frac{1}{w - 1} d w = ln ∣ w - 1 ∣

= \frac{1}{2} \cdot \frac{1}{α} (- ln ∣ w ∣ + ln ∣ w - 1 ∣)

Substitute back

= \frac{1}{2} \cdot \frac{1}{α} (- ln \frac{x ^{2} + α}{α} + ln \frac{x ^{2} + α}{α} - 1)

Simplify

\frac{1}{2} \cdot \frac{1}{α} (- ln \frac{x ^{2} + α}{α} + ln \frac{x ^{2} + α}{α} - 1) : \frac{1}{2 α} (- ln (\frac{x ^{2} + α}{∣ α ∣}) + ln \frac{x ^{2} + α}{α} - 1)

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

Add a constant to the solution

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

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What is the integral of 1/(x(x^2+α)) ?
The integral of 1/(x(x^2+α)) is 1/(2α)(-ln((|x^2+α|}{|α|})+ln|\frac{x^2+α)/α-1|)+C