Steps of {∫ [ln (LNX)] / X} DX

Steps of {∫ [ln (LNX)] / X} DX

I=∫ [ln(lnx)]/x dx
let y = lnx
dy = 1/x dx
I= ∫ ln y dy
= 1/y + C
= 1/lnx + C

x+ln(x^2)/x dx lnx/x(1+lnx)^/1/2 dx 2/3 (1+ln)^2/3-2(1+lnx)^1/2+c

∫x+ln(x^2)/x dx
∫ xdx + ∫ ln (X & # 178;) / xdx = x & # 178; / 2 + ∫ (1 / 2) ln (X & # 178;) DLN (X & # 178;) = x & # 178;) / 2 + (1 / 4) (LN (x & # 178;) &# 178; + C (C is constant)
∫lnx/x(1+lnx)^/1/2 dx
=∫ LNX (1 + LNX) ^ (1 / 2) d (LNX + 1) = LNX × (2 / 3) × (1 + LNX) ^ (3 / 2) - ∫ (2 / 3) (1 + LNX) ^ (3 / 2) dlnx = LNX × (2 / 3) × (1 + LNX) ^ (3 / 2) - ∫ (2 / 3) (1 + LNX) ^ (3 / 2) d (LNX + 1) = LNX × (2 / 3) × (1 + LNX) ^ (3 / 2) - (4 / 15) (1 + LNX) ^ (5 / 2) + C (C is a constant
）
Answer 2 / 3 (1 + LN) ^ 2 / 3-2 (1 + LNX) ^ 1 / 2 + C

Let f (x2-1) = lnx2x2 − 2, and f [φ (x)] = LNX, find ∫ φ (x) DX

T = x2-1, then x2 = t + 1; so: F (x2-1) = lnx2x2 − 2 = f (T) = LNT + 1T − 1, that is: F (x) = LNX + 1x − 1, so: F (φ (x)) = ln φ (x) + 1 φ (x) − 1 = LNX; so: φ (x) + 1 φ (x) − 1 = x; the solution is: φ (x) = x + 1x − 1; ∫ φ (x) DX = ∫ x + 1x − 1dx = ∫ X

If f · (LNX) = (LN (1 + x)) / X is known, then ∫ f (x) DX=

f（lnx）=(ln(1+x))/x
lnx=t
x=e^t
f(lnx)=f(t)=ln(1+e^t)/e^t
∫f(x)dx
=∫ln(1+e^x)/e^xdx
=∫ln(1+e^x)de^(-x)
=e^(-x)ln(1+e^x)-∫e^(-x)*1/(1+e^x)*e^xdx
=e^(-x)ln(1+e^x)-∫1/(1+e^x)dx
=e^(-x)ln(1+e^x)-∫e^x/(e^x+e^2x)dx
=e^(-x)ln(1+e^x)-∫1/(e^x+e^2x)de^x
=e^(-x)ln(1+e^x)-arctan[(e^x+1/2)/(√3/2)]+C

∫[ln(x+1)-lnx]/x(x+1) dx

Seeking the large of higher number & #128552; ∫ [- 1,1] (2x & #178; + xcosx) / (1 + √ 1-x & #178;) DX

It is known that the domain of F (x) is the domain of (0,1 / 2) to find f (arcsinx)

0 to 1 / 6

Let the domain of F (U) be 0

①∵0

Function f (x) = (√ 9-x ^ 2) / ln (x + 2), then the domain of F (LNX) is

Function f (x) = (√ 9-x ^ 2) / ln (x + 2)
f(lnx)=√[9-(lnx)^2]/ln[(lnx)+2)]
-2

Let arcsinx = t, x = Sint, why is x = Sint here?

Because sin (arcsinx) = X