What is the integral of 1/((x^2-4)^{1/2)} ?

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

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integral of 1/((x^2-4)^{1/2)}

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x}{( x ^{2} - 4 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

= \frac{x}{( x ^{2} - 4 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

\int - \frac{x ^{2}}{( x ^{2} - 4 ) ^{\frac{3}{2}}} d x = \frac{x}{( x ^{2} - 4 ) ^{\frac{1}{2}}} - ln (\frac{1}{2} (x^{2} - 4)^{\frac{1}{2}} + x)

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

Simplify

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

Add a constant to the solution

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

(1 + e^{x}) y y^{^{'}} = e^{x}, y (0) = 1

f (x) = 3 x^{2} - 8 x - x

x \to - 3 lim (x^{3} - 4 x^{2} - 7)

n = 1 \sum \infty \frac{n - 9}{n + 9}

\frac{( 11 s ^{2} - 2 s + 5 )}{( s - 2 ) ( 2 s - 1 ) ( s + 1 )}

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