What is the integral of 8/(sqrt(x^2-36)) ?

The integral of 8/(sqrt(x^2-36)) is 8ln(1/6 |sqrt(x^2-36)+x|)+C

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integral of 8/(sqrt(x^2-36))

\int \frac{8}{x ^{2} - 36} d x

8 ln (\frac{1}{6} x^{2} - 36 + x) + C

Solution steps

\int \frac{8}{x ^{2} - 36} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} - 36} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 8 \cdot \int sec (u) d u

Use the common integral:

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

= 8 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{6} x)

= 8 ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x))

Simplify

8 ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)) : 8 ln (\frac{1}{6} x^{2} - 36 + x)

= 8 ln (\frac{1}{6} x^{2} - 36 + x)

Add a constant to the solution

= 8 ln (\frac{1}{6} x^{2} - 36 + x) + C

\frac{d}{d x} (e^{6} + ln (9))

f (x)^{8} x^{2} + 4 x - 7, x = - 3

\int sec^{2} (x t a) n^{5} x d x

(\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) (y + 3 y^{3})

x \to 3 + lim (18 ln (x - 3))

What is the integral of 8/(sqrt(x^2-36)) ?
The integral of 8/(sqrt(x^2-36)) is 8ln(1/6 |sqrt(x^2-36)+x|)+C