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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

x^{2} \frac{d y}{d x} - y x = 0

n = 0 \sum \infty e^{- 2 n}

x \to - 4 - lim (5 x + 20 - 8)

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

\frac{\partial}{\partial y} (x + e^{z} + y)

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