integrate: (square root of (y^2 -121))/y dy, y>11. I got

(12(square root of (y^2-121))/12 -sec^-1 (y/12) +C

but it's not right. I'm not sure where I am going wrong.

To integrate the given expression ∫(√(y^2 - 121))/y dy, you're on the right track by employing the technique of trigonometric substitution. However, there seems to be a mistake in your calculation. Let's step through the correct approach:

1. Start by identifying a suitable trigonometric substitution. In this case, we can let y = 11sec(θ), so that dy = 11sec(θ)tan(θ) dθ.

2. Next, express the expression inside the square root using the trigonometric substitution: y^2 - 121 = 121sec^2(θ) - 121 = 121(tan^2(θ)).

3. Rewrite the original integral in terms of θ: ∫(√(121tan^2(θ)))/(11sec(θ)) * 11sec(θ)tan(θ) dθ.

Simplifying, we now have: ∫(11tan(θ))/sec(θ) * sec(θ)tan(θ) dθ.

4. Cancel out the sec(θ) terms to obtain: ∫(11tan^2(θ)) dθ.

5. Employ the trigonometric identity for tan^2(θ): tan^2(θ) = sec^2(θ) - 1.

6. Substitute this identity and rewrite the integral as follows: ∫(11(sec^2(θ) - 1)) dθ.

Expanding, we have: ∫(11sec^2(θ)) dθ - ∫(11) dθ.

7. Integrate each term: 11∫sec^2(θ) dθ - 11∫1 dθ.

∫sec^2(θ) dθ is a standard integral, which equals tan(θ).

The integral ∫1 dθ is simply θ.

8. Replace θ with the inverse trigonometric substitution y/11 in the final answer:

11tan(θ) - 11θ + C = 11tan(sec^(-1)(y/11)) - 11(sec^(-1)(y/11)) + C.

Recall that tan(sec^(-1)(x)) is equivalent to √(x^2 + 1).

Therefore, the final answer is:

11√((y/11)^2 + 1) - 11(sec^(-1)(y/11)) + C.

This should be the correct result for integrating the given expression.