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

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

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

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

\frac{1}{4} arcsec (\frac{1}{4} (x + 2)) + C

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= \int \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 1 \cdot v

Substitute back

= \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} (x + 2))

Simplify

= \frac{1}{4} arcsec (\frac{1}{4} (x + 2))

Add a constant to the solution

= \frac{1}{4} arcsec (\frac{1}{4} (x + 2)) + C

\frac{1}{1 + x ^{2}}

x \to 0 - lim (6 x - 1)

\frac{s}{( s + 1 ) ( s ^{2} - s + 1 )}

x \to 0 lim (\frac{x + ∣ x ∣}{x})

\frac{d}{d x} (5 x + 7)

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