What is the integral of-((x+1))/(x^2-1) ?

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

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integral of-((x+1))/(x^2-1)

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

- ln ∣ x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Factor

\frac{x + 1}{x ^{2} - 1}

= - \int \frac{x + 1}{( x + 1 ) ( x - 1 )} d x

Apply u-substitution

: \int \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = x - 1

= - ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + C

\int 9 cos (5 x) d x

\frac{d}{d x} (8 e^{6 x})

y^{^{'}} - 5 y = 2 e^{(5 x)}

\int_{0}^{\frac{π}{2}} 5 sin^{2} (2 x) d x

f (x) = 5 - 2 x^{2}, at x = - 3, 13

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