What is the integral of 8/(5xsqrt(25x^2-25)) ?

The integral of 8/(5xsqrt(25x^2-25)) is 8/25 arctan(sqrt(x^2-1))+C

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integral of 8/(5xsqrt(25x^2-25))

\int \frac{8}{5 x 25 x ^{2} - 25} d x

\frac{8}{25} arctan (x^{2} - 1) + C

Solution steps

\int \frac{8}{5 x 25 x ^{2} - 25} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{8}{5} \cdot \int \frac{1}{x 25 x ^{2} - 25} d x

Apply Integration By Parts

: \frac{1}{5 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{5 x ( x ^{2} - 1 ) x ^{2} - 1} d x

= \frac{8}{5} (\frac{1}{5 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{5 x ( x ^{2} - 1 ) x ^{2} - 1} d x)

\int - \frac{2 x ^{2} - 1}{5 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{5} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

= \frac{8}{5} (\frac{1}{5 x ^{2} - 1} - (- \frac{1}{5} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))))

Simplify

\frac{8}{5} (\frac{1}{5 x ^{2} - 1} - (- \frac{1}{5} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1)))) : \frac{8}{25} arctan (x^{2} - 1)

= \frac{8}{25} arctan (x^{2} - 1)

Add a constant to the solution

= \frac{8}{25} arctan (x^{2} - 1) + C

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What is the integral of 8/(5xsqrt(25x^2-25)) ?
The integral of 8/(5xsqrt(25x^2-25)) is 8/25 arctan(sqrt(x^2-1))+C