What is the integral of 1/(sqrt(x^2+6x+18)) ?

The integral of 1/(sqrt(x^2+6x+18)) is ln(1/3 |x+3+sqrt(x^2+6x+18)|)+C

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integral of 1/(sqrt(x^2+6x+18))

\int \frac{1}{x ^{2} + 6 x + 18} d x

ln (\frac{1}{3} x + 3 + x^{2} + 6 x + 18) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 18} d x

Complete the square

x^{2} + 6 x + 18 : (x + 3)^{2} + 9

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

ln tan (arctan (\frac{1}{3} (x + 3))) + sec (arctan (\frac{1}{3} (x + 3))) : ln (\frac{1}{3} x + 3 + x^{2} + 6 x + 18)

= ln (\frac{1}{3} x + 3 + x^{2} + 6 x + 18)

Add a constant to the solution

= ln (\frac{1}{3} x + 3 + x^{2} + 6 x + 18) + C

x \to 1 lim (x - 4)

\int \frac{( arctan ( x ) ) ^{5}}{x ^{2} + 1} d x

\frac{d}{d x} ((x^{3} - 8)^{\frac{2}{3}})

x \to 3 - lim (x^{2} - 8)

\frac{\partial}{\partial x} (- e^{x} cos (y))

What is the integral of 1/(sqrt(x^2+6x+18)) ?
The integral of 1/(sqrt(x^2+6x+18)) is ln(1/3 |x+3+sqrt(x^2+6x+18)|)+C