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

\int \frac{1}{sec ( y ) tan ( y )} d y

y^{^{'}} + 2 y = e^{3 x}

\frac{d}{d x} (5 tan (2 x^{3}))

\int b x d x

n = 1 \sum \infty 9 (\frac{1}{10})^{n}

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