What is the integral of 1/(sqrt(x^2+4x-12)) ?

The integral of 1/(sqrt(x^2+4x-12)) is ln(1/4 |sqrt(x^2+4x-12)+x+2|)+C

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integral of 1/(sqrt(x^2+4x-12))

\int \frac{1}{x ^{2} + 4 x - 12} d x

ln (\frac{1}{4} x^{2} + 4 x - 12 + x + 2) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x - 12} d x

Complete the square

x^{2} + 4 x - 12 : (x + 2)^{2} - 16

= \int \frac{1}{( x + 2 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 16} d u

= \int \frac{1}{u ^{2} - 16} 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 (arcsec (\frac{1}{4} (x + 2))) + sec (arcsec (\frac{1}{4} (x + 2)))

Simplify

ln tan (arcsec (\frac{1}{4} (x + 2))) + sec (arcsec (\frac{1}{4} (x + 2))) : ln (\frac{1}{4} x^{2} + 4 x - 12 + x + 2)

= ln (\frac{1}{4} x^{2} + 4 x - 12 + x + 2)

Add a constant to the solution

= ln (\frac{1}{4} x^{2} + 4 x - 12 + x + 2) + C

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 9

cos (e^{- 4 θ^{5}})

\frac{d}{d x} (\frac{9 x ^{2} - 6}{x ^{2}})

\frac{\partial}{\partial x} (\frac{4}{1 + x ^{2}})

\frac{d}{d x} (1 - x y cos (π y))

What is the integral of 1/(sqrt(x^2+4x-12)) ?
The integral of 1/(sqrt(x^2+4x-12)) is ln(1/4 |sqrt(x^2+4x-12)+x+2|)+C