What is the integral of 1/(sqrt(9+x^2)) ?

The integral of 1/(sqrt(9+x^2)) is ln(1/3 |x+sqrt(9+x^2)|)+C

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integral of 1/(sqrt(9+x^2))

\int \frac{1}{9 + x ^{2}} d x

ln (\frac{1}{3} x + 9 + x^{2}) + C

Solution steps

\int \frac{1}{9 + 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}{3} x)

= ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

Simplify

ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) : ln (\frac{1}{3} x + 9 + x^{2})

= ln (\frac{1}{3} x + 9 + x^{2})

Add a constant to the solution

= ln (\frac{1}{3} x + 9 + x^{2}) + C

y = \frac{- 5}{3 x}

f (x) = 2 x^{2} - 4 x - 5, at x = 2

\int sin (e^{x}) e^{x} d x

\frac{d y}{d x} = \frac{3 y ^{2} + 10 x ^{2}}{x y}

\frac{d}{d x} (7 x - 5)

What is the integral of 1/(sqrt(9+x^2)) ?
The integral of 1/(sqrt(9+x^2)) is ln(1/3 |x+sqrt(9+x^2)|)+C