What is the integral of 9/(sqrt(3+x^2)) ?

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

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integral of 9/(sqrt(3+x^2))

\int \frac{9}{3 + x ^{2}} d x

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

Solution steps

\int \frac{9}{3 + x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{3 + x ^{2}} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 9 \cdot \int sec (u) d u

Use the common integral:

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

= 9 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= 9 ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

Simplify

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

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

Add a constant to the solution

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

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