What is the integral of 1/(xsqrt(16x^2-1)) ?

The integral of 1/(xsqrt(16x^2-1)) is arctan(sqrt(16x^2-1))+C

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integral of 1/(xsqrt(16x^2-1))

\int \frac{1}{x 16 x ^{2} - 1} d x

arctan (16 x^{2} - 1) + C

Solution steps

\int \frac{1}{x 16 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 = 16 x^{2} - 1

= arctan (16 x^{2} - 1)

Add a constant to the solution

= arctan (16 x^{2} - 1) + C

f (x) = \frac{t}{( t - 7 ) ^{2}}

\frac{d}{d x} (ln (e^{2 x} + 1))

x \to \infty lim ((1 + \frac{1}{x})^{x + 3})

y^{^{''}} = - k y

- sin (x^{2}) (2 x)

What is the integral of 1/(xsqrt(16x^2-1)) ?
The integral of 1/(xsqrt(16x^2-1)) is arctan(sqrt(16x^2-1))+C