What is the integral of 1/((x^2-25)^{1/2)} ?

The integral of 1/((x^2-25)^{1/2)} is ln(1/5 |(x^2-25)^{1/2}+x|)+C

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integral of 1/((x^2-25)^{1/2)}

\int \frac{1}{( x ^{2} - 25 ) ^{\frac{1}{2}}} d x

ln (\frac{1}{5} (x^{2} - 25)^{\frac{1}{2}} + x) + C

Solution steps

\int \frac{1}{( x ^{2} - 25 ) ^{\frac{1}{2}}} d x

Apply Integration By Parts

: \frac{x}{( x ^{2} - 25 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 25 ) ^{\frac{3}{2}}} d x

= \frac{x}{( x ^{2} - 25 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 25 ) ^{\frac{3}{2}}} d x

\int - \frac{x ^{2}}{( x ^{2} - 25 ) ^{\frac{3}{2}}} d x = \frac{x}{( x ^{2} - 25 ) ^{\frac{1}{2}}} - ln (\frac{1}{5} (x^{2} - 25)^{\frac{1}{2}} + x)

= \frac{x}{( x ^{2} - 25 ) ^{\frac{1}{2}}} - (\frac{x}{( x ^{2} - 25 ) ^{\frac{1}{2}}} - ln (\frac{1}{5} (x^{2} - 25)^{\frac{1}{2}} + x))

Simplify

= ln (\frac{1}{5} (x^{2} - 25)^{\frac{1}{2}} + x)

Add a constant to the solution

= ln (\frac{1}{5} (x^{2} - 25)^{\frac{1}{2}} + x) + C

f (x) = - x^{3}, at x = 2

x \to 0 lim (2 x e^{2 x} + k x)

y^{^{'}} = 3 x + y, y (0) = 6

y = ln (e^{- x} + x e^{- x})

\frac{\partial}{\partial x} (- \frac{c x}{x ^{2} + y ^{2}})

What is the integral of 1/((x^2-25)^{1/2)} ?
The integral of 1/((x^2-25)^{1/2)} is ln(1/5 |(x^2-25)^{1/2}+x|)+C