Find integrate (1 / (xsqrt (1 + (lnx) ^2)),1, e, x) ... ?

Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as

(x² + x + 3 + x) / (x² + x + 3) = 1 + x / (x² + x + 3).

Next, complete the square.

x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2) ² + (√ (11) / 2) ²

Also, for the x in the numerator

x = x + 1/2 - 1/2.

So

(x² + 2x + 3) / (x² + x + 3) = 1 + (x + 1/2) / [ (x + 1/2) ² + (√ (11) / 2) ²] - 1/2/[ (x + 1/2) ² + (√ (11) / 2) ²].

Integrate term by term to get

∫ (x² + 2x + 3) / (x² + x + 3) dx = x + (1/2) ln (x² + x + 3) - (1/√ (11)) arctan (2 (x + 1/2) / √ (11)) + C

b) Use the fact that ln (x) = 2 ln√ (x). Then put u = √ (x), du = 1/[2√ (x) ] dx.

∫ ln (x) / √ (x) dx = 4 ∫ ln u du = 4 u ln (u) - u + C = 4√ (x) ln√ (x) - √ (x) + C

= 2 √ (x) ln (x) - √ (x) + C.

c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x.

∫ 1 / (e^ (-x) + e^x) dx = ∫ e^x / (1 + e^ (2x)) dx = ∫ du / (1 + u²) = arctan (u) + C

= arctan (e^x) + C.