Determine the indefinite integral of y=x/square root(x^2+9), using substitution.
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We have to find the integral of y= x / sqrt(x^2+9)
Int [ x/ sqrt (x^2 + 9) dx]
let x^2 + 9 = u, du / 2 = x dx
=> (1/2)*Int [ 1/ sqrt u du]
=> (1/2)*Int [ u^(-1/2) du]
=> (1/2) u^(1/2) / (1/2)
=> u^(1/2)
substitute u = x^2 + 9
=> sqrt (x^2 + 9) + C
The required integral is sqrt (x^2 + 9) + C
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We'll substitute the radicand x^2+9 by t.
x^2+9 = t
We'll differentiate both sides:
d(x^2+9)/dx = dt
2xdx = dt => xdx = dt/2
We'll solve the indefinite integral in t:
Int x dx/sqrt(x^2+9) = Int dt/2sqrt t = (1/2)*Int dt/t^1/2
(1/2)*Int dt/t^1/2 = (1/2)*Int t^(-1/2)*dt
(1/2)*Int t^(-1/2)*dt = (1/2)*t^(-1/2 + 1)/(-1/2 + 1) + C
(1/2)*Int t^(-1/2)*dt = (1/2)*2t^(1/2) + C
(1/2)*Int t^(-1/2)*dt = t^(1/2) + C
(1/2)*Int t^(-1/2)*dt = sqrt t + C
The indefinite integral of the given function is: Int xdx/sqrt(x^2+9) = sqrt(x^2 + 9) + C.
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