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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

e^{- 6 x}

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

h (x) = lo g_{9} (x)

\frac{d}{d x} (7 cos (x) + 3 e^{x})

y = 1 + cot^{2} (x)

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