Find, using substitution, the integral of:
e^x /sqrt[e^(2x) -1] dx
let u = e^x
then u^2 = e^(2x)
and du/dx = e^x
dx = du/e^x = du/u
ʃ e^x /sqrt[e^(2x) -1] dx
= ʃ u/√(u^2 - 1) du/u
= ʃ 1/√(u^2 - 1) du
= ln (√(u^2 - 1) + u) , from my old integration formulas
= ln((√(e^(2x) - 1) + e^x) + C
I've just found that integration formula (which I'd forgotten about). No wonder I kept getting stuck! Thank you.
To solve this integral using substitution, let's start by making a substitution. Let u = e^x, so du = e^x dx. We can rewrite the integral in terms of u:
∫ (e^x) / √[(e^(2x) - 1)] dx = ∫ (1/u) / √[(u^2 - 1)] du
Next, we need to simplify the expression inside the square root. Notice that u^2 - 1 = (u - 1)(u + 1). So, we can rewrite the integral as:
∫ (1/u) / √[(u - 1)(u + 1)] du
Now, let's separate the fraction inside the integral into two parts:
∫ (1/u) * [(1/√(u - 1))(1/√(u + 1))] du
Now, we can rewrite the integral as a product of two separate integrals:
∫ (1/√(u - 1)) du * ∫ (1/√(u + 1)) du
Let's now solve each of these integrals individually.
For the first integral, u = u - 1 + 1, which means we can rewrite it as:
∫ (1/√(u - 1)) du = ∫ (1/√(u - 1 + 1)) du = ∫ (1/√(u - 1 + 1)) du
Now, let's make a substitution for u - 1 + 1:
Let w = u - 1 + 1, so dw = du. Therefore, we can rewrite the integral as:
∫ (1/√w) dw = 2√w
Now, for the second integral, u = u + 1 - 1, which means we can rewrite it as:
∫ (1/√(u + 1)) du = ∫ (1/√(u + 1 - 1)) du = ∫ (1/√(u + 1 - 1)) du
Now, let's make a substitution for u + 1 - 1:
Let v = u + 1 - 1, so dv = du. Therefore, we can rewrite the integral as:
∫ (1/√v) dv = 2√v
Now, let's bring back the original variables:
2√w = 2√(u - 1)
2√v = 2√(u + 1)
Finally, we can write the complete solution:
∫ (e^x) / √[(e^(2x) - 1)] dx = 2√(e^x - 1) + 2√(e^x + 1) + C,
where C is the constant of integration.