What is the integral of 1/(sqrt(x^2-361)) ?

The integral of 1/(sqrt(x^2-361)) is ln(1/19 |sqrt(x^2-361)+x|)+C

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integral of 1/(sqrt(x^2-361))

\int \frac{1}{x ^{2} - 361} d x

ln (\frac{1}{19} x^{2} - 361 + x) + C

Solution steps

\int \frac{1}{x ^{2} - 361} 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}{19} x)

= ln tan (arcsec (\frac{1}{19} x)) + sec (arcsec (\frac{1}{19} x))

Simplify

ln tan (arcsec (\frac{1}{19} x)) + sec (arcsec (\frac{1}{19} x)) : ln (\frac{1}{19} x^{2} - 361 + x)

= ln (\frac{1}{19} x^{2} - 361 + x)

Add a constant to the solution

= ln (\frac{1}{19} x^{2} - 361 + x) + C

x \to 0 lim (\frac{x}{- 3 x ^{3} + 2})

x \to \infty lim (e^{- x})

\int (x^{3} + 3 x^{2}) d x

\int \frac{( ln ( x ) ) ^{15}}{x} d x

y = x^{3} - 2 x^{2} + 5

What is the integral of 1/(sqrt(x^2-361)) ?
The integral of 1/(sqrt(x^2-361)) is ln(1/19 |sqrt(x^2-361)+x|)+C