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

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

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

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

2 ln x^{2} - 1 + x + C

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= 2 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (x)

= 2 ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣

Simplify

2 ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ : 2 ln x^{2} - 1 + x

= 2 ln x^{2} - 1 + x

Add a constant to the solution

= 2 ln x^{2} - 1 + x + C

\int 3 x x^{2} - 3 d x

\int \frac{e ^{2 x}}{e ^{2 x} - 8} d x

\int \frac{2 x ^{2} - 4 x - 3}{x ^{2}} d x

tan^{2} (x) sec^{2} (x)

\int e^{t x} x d x

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