What is the integral of 1/(m^3+m) ?

The integral of 1/(m^3+m) is ln|m|-1/2 ln|m^2+1|+C

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integral of 1/(m^3+m)

\int \frac{1}{m ^{3} + m} d m

ln ∣ m ∣ - \frac{1}{2} ln m^{2} + 1 + C

Solution steps

\int \frac{1}{m ^{3} + m} d m

Take the partial fraction of

\frac{1}{m ^{3} + m} : \frac{1}{m} - \frac{m}{m ^{2} + 1}

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

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}{m} d m - \int \frac{m}{m ^{2} + 1} d m

\int \frac{1}{m} d m = ln ∣ m ∣

\int \frac{m}{m ^{2} + 1} d m = \frac{1}{2} ln m^{2} + 1

= ln ∣ m ∣ - \frac{1}{2} ln m^{2} + 1

Add a constant to the solution

= ln ∣ m ∣ - \frac{1}{2} ln m^{2} + 1 + C

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What is the integral of 1/(m^3+m) ?
The integral of 1/(m^3+m) is ln|m|-1/2 ln|m^2+1|+C