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

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

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

integral

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

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

Solution steps

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

Apply Trigonometric Substitution

= \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

in t e g r a l x^{4} \cdot n^{- 1} \cdot 1 - x^{2}

in t e g r a l y = 2 e^{4} x + 3

in t e g r a l - 3 sec (x) \cdot sec^{3} (x)

in t e g r a l - sec^{2} (x)

in t e g r a l - 3 sec (x) \cdot (sec (x))^{3}

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