# 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**

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.**