What is the integral of 1/(xsqrt(9x^2-1)) ?

The integral of 1/(xsqrt(9x^2-1)) is arctan(sqrt(9x^2-1))+C

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integral of 1/(xsqrt(9x^2-1))

\int \frac{1}{x 9 x ^{2} - 1} d x

arctan (9 x^{2} - 1) + C

Solution steps

\int \frac{1}{x 9 x ^{2} - 1} d x

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 = 9 x^{2} - 1

= arctan (9 x^{2} - 1)

Add a constant to the solution

= arctan (9 x^{2} - 1) + C

x^{^{'}} = \frac{1}{1 + t ^{2}}

f (x) = 4 - 2 x^{2}, at x = 5

(\frac{- 2 x - 1}{- 4 x + 2})^{5}

\frac{\partial}{\partial y} (x^{2} + x y^{4})

\int_{1}^{x} \frac{1}{t} d t

What is the integral of 1/(xsqrt(9x^2-1)) ?
The integral of 1/(xsqrt(9x^2-1)) is arctan(sqrt(9x^2-1))+C