What is the integral of 1/(xsqrt(256x^2-256)) ?

The integral of 1/(xsqrt(256x^2-256)) is 1/16 arctan(sqrt(x^2-1))+C

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integral of 1/(xsqrt(256x^2-256))

\int \frac{1}{x 256 x ^{2} - 256} d x

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

Solution steps

\int \frac{1}{x 256 x ^{2} - 256} d x

Apply Integration By Parts

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

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

\int - \frac{2 x ^{2} - 1}{16 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{16} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

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

Simplify

\frac{1}{16 x ^{2} - 1} - (- \frac{1}{16} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))) : \frac{1}{16} arctan (x^{2} - 1)

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

Add a constant to the solution

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

x \to 1 - lim ((x - 1)^{2} f (x))

cot (x^{7})

x \to 2 lim (x 8 - x^{2})

\int \frac{1}{16 x ^{2} - 25} d x

\frac{d}{d x} (π)

What is the integral of 1/(xsqrt(256x^2-256)) ?
The integral of 1/(xsqrt(256x^2-256)) is 1/16 arctan(sqrt(x^2-1))+C