What is the integral of 3/((u^2-2u)) ?

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

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integral of 3/((u^2-2u))

integral

\frac{3}{( u ^{2} - 2 u )}

- 3 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 2 ∣) + C

Solution steps

\int \frac{3}{u ^{2} - 2 u} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{u ^{2} - 2 u} d u

Complete the square

u^{2} - 2 u : (u - 1)^{2} - 1

= 3 \cdot \int \frac{1}{( u - 1 ) ^{2} - 1} d u

Apply u-substitution

: \int \frac{1}{v ^{2} - 1} d v

= 3 \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 3 \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 3 (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 3 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

v = u - 1

= 3 (- (\frac{ln ∣ u - 1 + 1 ∣}{2} - \frac{ln ∣ u - 1 - 1 ∣}{2}))

Simplify

3 (- (\frac{ln ∣ u - 1 + 1 ∣}{2} - \frac{ln ∣ u - 1 - 1 ∣}{2})) : - 3 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 2 ∣)

= - 3 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 2 ∣)

Add a constant to the solution

= - 3 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 2 ∣) + C

\frac{3}{x + 1}

f (\frac{tan ( 1 )}{x})^{2}

3 x^{- 4}

\frac{1}{( x ^{2} + c )}

(cos (x))^{3}

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