What is the integral of 2/(xsqrt(x^2-4)) ?

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

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integral of 2/(xsqrt(x^2-4))

\int \frac{2}{x x ^{2} - 4} d x

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

Solution steps

\int \frac{2}{x x ^{2} - 4} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= 2 \cdot \int \frac{1}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{2} x)

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

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} x) : arcsec (\frac{1}{2} x)

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

Add a constant to the solution

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

\int_{3}^{4} - lo g_{9} (x) d x

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

\frac{\partial}{\partial y} (x y e^{9 y})

y \to + 0 lim (\frac{y}{sin ( 5 y )})

g (x) = x^{\frac{2}{3}}

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