What is the integral of 1/(m^3+m^2) ?

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

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integral of 1/(m^3+m^2)

\int \frac{1}{m ^{3} + m ^{2}} d m

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

Solution steps

\int \frac{1}{m ^{3} + m ^{2}} d m

Take the partial fraction of

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

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

\int \frac{1}{m} d m = ln ∣ m ∣

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

\int \frac{1}{m + 1} d m = ln ∣ m + 1 ∣

= - ln ∣ m ∣ - \frac{1}{m} + ln ∣ m + 1 ∣

Add a constant to the solution

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

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