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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 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 \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

v = (u + 1) (u - 1)

= \frac{1}{2} ln ∣ (u + 1) (u - 1) ∣

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{3 x^{3} - x})

x^{4} - x^{3} + 2

\frac{d}{d x} (\frac{1}{2} x^{2} e^{- 2 x})

\frac{2 x + 9}{x}

\int \frac{y ^{3}}{( 1 - 2 y ^{4} ) ^{5}} d y

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