What is the integral of 3/(sqrt(x^2-1)) ?

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

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integral of 3/(sqrt(x^2-1))

\int \frac{3}{x ^{2} - 1} d x

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

Solution steps

\int \frac{3}{x ^{2} - 1} d x

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= 3 \cdot \int sec (u) d u

Use the common integral:

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

= 3 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (x)

= 3 ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣

Simplify

3 ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ : 3 ln x^{2} - 1 + x

= 3 ln x^{2} - 1 + x

Add a constant to the solution

= 3 ln x^{2} - 1 + x + C

\int 3 x e^{\frac{2 x}{3}} d x

\frac{d}{d x} (4 - 7 x)

\frac{d y}{d x} = 100 x - x y

e^{7 e^{x}}

\int_{- \infty}^{0} 5^{r} d r

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