What is the integral of 1/(sqrt(x^2+12x+48)) ?

The integral of 1/(sqrt(x^2+12x+48)) is ln(1/(2sqrt(3))|x+6+sqrt(x^2+12x+48)|)+C

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integral of 1/(sqrt(x^2+12x+48))

\int \frac{1}{x ^{2} + 12 x + 48} d x

ln (\frac{1}{2 3} x + 6 + x^{2} + 12 x + 48) + C

Solution steps

\int \frac{1}{x ^{2} + 12 x + 48} d x

Complete the square

x^{2} + 12 x + 48 : (x + 6)^{2} + 12

= \int \frac{1}{( x + 6 ) ^{2} + 12} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 12} d u

= \int \frac{1}{u ^{2} + 12} 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}{2 3} (x + 6))) + sec (arctan (\frac{1}{2 3} (x + 6)))

Simplify

ln tan (arctan (\frac{1}{2 3} (x + 6))) + sec (arctan (\frac{1}{2 3} (x + 6))) : ln (\frac{1}{2 3} x + 6 + x^{2} + 12 x + 48)

= ln (\frac{1}{2 3} x + 6 + x^{2} + 12 x + 48)

Add a constant to the solution

= ln (\frac{1}{2 3} x + 6 + x^{2} + 12 x + 48) + C

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What is the integral of 1/(sqrt(x^2+12x+48)) ?
The integral of 1/(sqrt(x^2+12x+48)) is ln(1/(2sqrt(3))|x+6+sqrt(x^2+12x+48)|)+C