What is the integral of 1/(u(u-1)) ?

The integral of 1/(u(u-1)) is -ln|u|+ln|u-1|+C

Popular >

integral of 1/(u(u-1))

\int \frac{1}{u ( u - 1 )} d u

- ln ∣ u ∣ + ln ∣ u - 1 ∣ + C

Solution steps

\int \frac{1}{u ( u - 1 )} d u

Take the partial fraction of

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

= \int - \frac{1}{u} + \frac{1}{u - 1} d u

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}{u} d u + \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - ln ∣ u ∣ + ln ∣ u - 1 ∣

Add a constant to the solution

= - ln ∣ u ∣ + ln ∣ u - 1 ∣ + C

\int \frac{1}{sin ^{2} ( 2 x ) cos ^{4} ( 2 x )} d x

\int \frac{x ^{2} - 5 x + 16}{( 2 - x ) ^{2} ( 2 x + 1 )} d x

\int e^{- x^{2} + 2 x} d x

\int \frac{1}{( 2 sin ( x ) - cos ( x ) + 5 )} d x

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

What is the integral of 1/(u(u-1)) ?
The integral of 1/(u(u-1)) is -ln|u|+ln|u-1|+C