How do I evaluate ∫01eex√dx+2∫0ee√ln(lnx)dx?

Let's evaluate the integrals one by one:

First Integral:

∫₀¹ e^x√dx

We can use integration by substitution with u = √x and du = 1/2√x dx, which gives:

∫₀¹ e^x√dx = 2∫₀¹ e^(u^2)du

This integral has no elementary function to express the solution in terms of the usual functions. It is known as the error function, which we can approximate using numerical methods. A common approximation is:

∫₀¹ e^(u^2)du ≈ 0.7468

Second Integral:

∫₀^e e^√ln(lnx)dx

Again, we can use integration by substitution with u = √ln(lnx), du = 1/(2√lnx ln(x)) dx, which gives:

∫₀^e e^√ln(lnx)dx = 2∫₀^√ln2 e^udu

= 2[e^u]₀^√ln2 = 2[e^√ln(ln2) - 1]

Therefore, the value of the given expression is:

∫₀¹ e^x√dx + 2∫₀^e e^√ln(lnx)dx

≈ 0.7468 + 2[e^√ln(ln2) - 1]

≈ 2.3725

Hence, the approximate value of the given expression is 2.3725.