What is the integral of 1/(xsqrt(25x^2-25)) ?

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

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integral of 1/(xsqrt(25x^2-25))

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

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

Solution steps

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

Simplify

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

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

Add a constant to the solution

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

x \to - 5.8 lim (ln (x^{2} - 9))

n = 1 \sum \infty \frac{n ^{2}}{n ^{3}}

\frac{1}{4} x

(e^{t} cos (t))^{^{'}}

\frac{\partial}{\partial y} (y^{2} - x)

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