Integrate [1/square root of(e^(2x)-1)]. I have to use u substitution. We are doing the integrals of inverse trig functions, but I cannot get it to work out!
To integrate the given expression ∫[1/√(e^(2x)-1)] dx using the u-substitution method, you can follow these steps:
1. Let's start by selecting a suitable substitution. In this case, we can let u = e^x. Therefore, du/dx = e^x, and dx = du/u.
2. Now, we need to express the current variables in terms of u. We can rewrite the expression e^(2x) as (e^x)^2, which is equal to u^2. Also, note that √(e^(2x)-1) becomes √(u^2-1) after substitution.
3. Substitute u and dx in terms of u into the original integral. The integral becomes:
∫ [1/√(e^(2x)-1)] dx = ∫ [1/√(u^2-1)] (du/u)
4. Simplify the expression:
∫ [1/√(u^2-1)] (du/u) = ∫ (1/u) * (1/√(u^2-1)) du
5. Now, we can separate the terms in the integral:
∫ (1/u) * (1/√(u^2-1)) du = ∫ 1/u^2 * 1/√(u^2-1) du
6. After separating the terms, we can evaluate each integral separately:
∫ 1/u^2 du = -1/u
∫ 1/√(u^2-1) du = arcsin(u)
7. Finally, substitute the original variable back into the expression:
∫ [1/√(e^(2x)-1)] dx = ∫ (1/u^2) * (1/√(u^2-1)) du = -1/u * arcsin(u) + C
Therefore, the final answer is -arcsin(e^x)/e^x + C, where C is the constant of integration.