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

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

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

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

- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

= - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

\frac{d}{d x} (- \frac{1}{64} sin (\frac{x}{8}))

\frac{d}{d t} (a cos (wt) + b sin (wt))

\int (e^{5 x}) (cos (- 2 x)) d x

\frac{d y}{d x} = y - x, y (0) = \frac{2}{3}

\int \frac{2 x + 1}{( x - 1 ) ( x + 3 )} d x

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