What is the integral of (sqrt(2x^2-4))/(x^3) ?

The integral of (sqrt(2x^2-4))/(x^3) is -(sqrt(2x^2-4))/(2x^2)+1/2 arctan(1/2 sqrt(2x^2-4))+C

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integral of (sqrt(2x^2-4))/(x^3)

\int \frac{2 x ^{2} - 4}{x ^{3}} d x

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

Solution steps

\int \frac{2 x ^{2} - 4}{x ^{3}} d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{17}{4}

\frac{d}{d x} (x^{\frac{3}{2}} + 4)

y = sin (cot (3 π x))

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 0

\frac{d}{d x} ((1 + e^{- x})^{x})

What is the integral of (sqrt(2x^2-4))/(x^3) ?
The integral of (sqrt(2x^2-4))/(x^3) is -(sqrt(2x^2-4))/(2x^2)+1/2 arctan(1/2 sqrt(2x^2-4))+C