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

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

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

\int \frac{1}{n ^{2} - 1}

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

Solution steps

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

Integral of a constant:

\int a d x = a x

= \frac{1}{n ^{2} - 1} x

Simplify

\frac{1}{n ^{2} - 1} x : \frac{x}{n ^{2} - 1}

= \frac{x}{n ^{2} - 1}

Add a constant to the solution

= \frac{x}{n ^{2} - 1} + C

y = 3 x^{3} - 3 x, (1, 0)

\frac{d y}{d x} = 2 y^{2} (x + 1)

\int cot (22 x) d x

\int \frac{1}{x ^{4} x ^{2} - 64} d x

\frac{\partial}{\partial x} (x^{l n (x)})

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