What is the antiderivative of y=x/square root(x^2-9) ?
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We have to find the antiderivative of y = x / sqrt ( x^2 - 9)
Let u = x^2 - 9
=> du / dx = 2x
=> x dx = du/2
Int [ x / sqrt ( x^2 - 9) dx]
=> Int [(1/2)*(1/ sqrt u) du]
=> Int [ (1/2)* u^ (-1/2) du]
=> (1/2)* u^( 1/2)/ (1/2) + C
=> u^(1/2) + C
replace u with x^2 - 9
=> sqrt (x^2 - 9) + C
The required integral is sqrt (x^2 - 9) + C
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We'll substitute x^2 - 9 = t
We'll differentiate both sides:
2xdx = dt
xdx = dt/2
We'll re-write the integral in t:
Int xdx/sqrt (x^2 - 9) = (1/2)*Int dt/sqrt t
(1/2)*Int dt/sqrt t = (1/2)*t^(-1/2 + 1)/(1 - 1/2) + C
(1/2)*Int dt/sqrt t = (1/2)*t^(1/2)/(1/2) + C
(1/2)*Int dt/sqrt t = t^(1/2) + C
(1/2)*Int dt/sqrt t = sqrt t + C
We'll substitute t by the original expression x^2 - 9:
The antiderivative of f(x) is: Int xdx/sqrt (x^2 - 9) = sqrt(x^2 - 9) + C
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