What is the integral of 1/(xsqrt(25x^2-4)) ?

The integral of 1/(xsqrt(25x^2-4)) is 1/2 arcsec(5/2 x)+C

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integral of 1/(xsqrt(25x^2-4))

\int \frac{1}{x 25 x ^{2} - 4} d x

\frac{1}{2} arcsec (\frac{5}{2} x) + C

Solution steps

\int \frac{1}{x 25 x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{5}{2} x)

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{5}{2} x)

Simplify

= \frac{1}{2} arcsec (\frac{5}{2} x)

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{5}{2} x) + C

\int \frac{e ^{y}}{1 - e ^{y}} d y

x \to 0 lim (\frac{0 x ^{4} + e ^{0 x}}{x ^{4}})

\frac{d}{d x} (x ln^{2} (x))

(y + \frac{3}{y}) (y^{2} - 5)

\frac{d}{d x} (sec^{n - 2} (x))

What is the integral of 1/(xsqrt(25x^2-4)) ?
The integral of 1/(xsqrt(25x^2-4)) is 1/2 arcsec(5/2 x)+C