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

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

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

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

arctan (n^{2} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= arctan (u)

Substitute back

u = n^{2} - 1

= arctan (n^{2} - 1)

Add a constant to the solution

= arctan (n^{2} - 1) + C

cos (e^{x^{5} c o s (x)})

\int_{6}^{8} \frac{96}{( x - 6 ) ^{3}} d x

\int e^{3 x} \cdot x d x

y = sin^{2} (x), y = sin^{3} (x), 0 \leq x \leq π

[x^{2} + y^{2}] d x = x y d y

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