What is the integral of 1/(sqrt(x^2+36)) ?

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

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integral of 1/(sqrt(x^2+36))

\int \frac{1}{x ^{2} + 36} d x

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

Solution steps

\int \frac{1}{x ^{2} + 36} 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 = arctan (\frac{1}{6} x)

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

Simplify

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

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

Add a constant to the solution

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

ln (3)

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

y = 4 x^{2}, y = x^{2} + 6

x \to 7 lim (x + 2)

\int x e^{x} ln (x - 1) d x

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