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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

: \int sec (u) d u

= 2 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{3} x)

= 2 ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{n}{1 + n ^{2}}

f (x) = \frac{2}{x ^{2}}, (7, \frac{2}{49})

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

\frac{d}{d y} (\frac{y}{y + \frac{b}{y}})

\int 2 x (x^{2} + 1)^{23} d x

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