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

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### 2 Answers

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**

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**