What is the integral of 4/((x^2-64)^{1/2)} ?

The integral of 4/((x^2-64)^{1/2)} is 4ln(1/8 |(x^2-64)^{1/2}+x|)+C

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integral of 4/((x^2-64)^{1/2)}

\int \frac{4}{( x ^{2} - 64 ) ^{\frac{1}{2}}} d x

4 ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x) + C

Solution steps

\int \frac{4}{( x ^{2} - 64 ) ^{\frac{1}{2}}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( x ^{2} - 64 ) ^{\frac{1}{2}}} d x

Apply Integration By Parts

: \frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x

= 4 (\frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} - \int - \frac{x ^{2}}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x)

\int - \frac{x ^{2}}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x = \frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} - ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x)

= 4 (\frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} - (\frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} - ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x)))

Simplify

= 4 ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x)

Add a constant to the solution

= 4 ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x) + C

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What is the integral of 4/((x^2-64)^{1/2)} ?
The integral of 4/((x^2-64)^{1/2)} is 4ln(1/8 |(x^2-64)^{1/2}+x|)+C