∫ln(2x)dx=

∫ln(2x)dx=

∫ ln(2x) dx
Partial integral
=xln(2x) - ∫ x*(1/x) dx
=xln(2x) - x + C
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1/2 ∫ln(1+2x)dx^2=?

1/2 ∫ln(1+2x)dx^2=1/2x^2ln(1+2x)-1/2 ∫x^2dln(1+2x)=1/2x^2ln(1+2x)- ∫x^2/(1+2x)dx=1/2x^2ln(1+2x)- ∫(x^2-1/4+1/4)/(1+2x)dx=1/2x^2ln(1+2x)- ∫[x/2-1/4+(1/4)/(1+2x)]dx=1/2x^2ln(1+2x)-x^2/4+x/4-1/8ln(1...

∫ln^2x / x（1+ln^2x） dx =∫(ln^2x +1-1)/(1+ln^2x)d(lnx) Where's x

∫ln^2x / x（1+ln^2x） dx
=∫(ln^2x +1-1)/(1+ln^2x)d(lnx)
=lnx-arctan(lnx)+c

Calculus ∫ 1 / (x ln ^ 2 x) DX

∫ 1 / (x Ln ² x ) dx= ∫ 1 / Ln ² X D LNX uses D LNX = 1 / x = ∫ 1 / LN ² X D LNX will use d (1 / x) = - 1 / X ² Note the minus sign. In fact, D (x ^ n) = NX ^ (n-1) when n = - 1 is the above = - ∫ 1 D (1 / LNX) =

Mathematical function problem; Y = f (x) domain [0,2], then what is the domain of function f (x + 1) + F (2x-1)

0 ≤ x + 1 ≤ 2 and 0 ≤ 2x-1 ≤ 2
-1 ≤ x ≤ 1 and 1 / 2 ≤ x ≤ 3 / 2
So 1 / 2 ≤ x ≤ 1

If f (x) domain [1,2] finds f (2x) domain I always don't understand how there is a relationship between the domain of F (2x) and the domain of F (x) If it is f (2x) domain [1,2], find f (x)

If you make y = 2x, then according to the meaning of the topic, the definition field of F (y) is [1,2], that is, 1