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

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

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

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

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

Solution steps

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

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\int_{0}^{4} x 19 - x^{2} d x

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\int_{0}^{10} x^{2} d x

\frac{d}{d x} (3 + e^{- 3 x} - 3 e^{- 3 x} x)

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