What is the integral of 1/(sqrt(x^2-49)) ?

The integral of 1/(sqrt(x^2-49)) is ln(1/7 |sqrt(x^2-49)+x|)+C

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integral of 1/(sqrt(x^2-49))

\int \frac{1}{x ^{2} - 49} d x

ln (\frac{1}{7} x^{2} - 49 + x) + C

Solution steps

\int \frac{1}{x ^{2} - 49} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{7} x)

= ln tan (arcsec (\frac{1}{7} x)) + sec (arcsec (\frac{1}{7} x))

Simplify

ln tan (arcsec (\frac{1}{7} x)) + sec (arcsec (\frac{1}{7} x)) : ln (\frac{1}{7} x^{2} - 49 + x)

= ln (\frac{1}{7} x^{2} - 49 + x)

Add a constant to the solution

= ln (\frac{1}{7} x^{2} - 49 + x) + C

x \to 2 - lim (x^{2} - 3)

\frac{d}{d x} (- \frac{6}{1 - 6 x})

y^{^{'}} \frac{1}{x} y = 0

\int x^{2} x^{3} + 11 d x

y = x^{2} - (sin (x))^{2}

What is the integral of 1/(sqrt(x^2-49)) ?
The integral of 1/(sqrt(x^2-49)) is ln(1/7 |sqrt(x^2-49)+x|)+C