What is the antiderivative of y=x/square root(x^2-9) ?

Expert Answers

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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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