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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= \int \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

h \to 0 lim (\frac{3 ^{h} - 1}{h})

f (x) = \frac{1}{x}, at x = 25

f (x) = (4 x - 3)^{\frac{1}{2}}

y^{^{''}} + y = - sin (2 x), y (0) = 0

\int \frac{1}{11 x - 5} d x

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