What is the integral of 1/(xsqrt(4x^2-1)) ?

The integral of 1/(xsqrt(4x^2-1)) is arctan(sqrt(4x^2-1))+C

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integral of 1/(xsqrt(4x^2-1))

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

arctan (4 x^{2} - 1) + C

Solution steps

\int \frac{1}{x 4 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 = 4 x^{2} - 1

= arctan (4 x^{2} - 1)

Add a constant to the solution

= arctan (4 x^{2} - 1) + C

x \to 1 + lim (arctan (ln (x - 1)))

n = 1 \sum \infty \frac{12}{- 5 ^{n}}

w e^{7 w}

f (x) = x^{- 15}

y^{^{'}} + \frac{1}{200 + 3 t} y = 12

What is the integral of 1/(xsqrt(4x^2-1)) ?
The integral of 1/(xsqrt(4x^2-1)) is arctan(sqrt(4x^2-1))+C