integral of ln(x)(sqrt(1-(ln(x))^2))/x
⚠Ⓘ ln(x)(sqrt(1-(ln(x))^2))/x is read as ln(x)*(sqrt(1-(ln(x))^2))/x: use (...)*(...) for multiplication, use (...)^(...) for a power
1) Let's integrate: ∫(ln(x)·1 -ln(x)2x)dx
For the math expression: ln(x)·1 -ln(x)2x, we know the derivative formula: ddx(an) = na(n -1), so we try differentiating with ^(m+1)
2) Let's differentiate: ddx((1 -ln(x)2)(12 + 1))
ddx((1 -ln(x)2)(1 + 22))
ddx((1 -ln(x)2)32)
Apply the differentiation rule: ddx(fn) = n·ddx(f)·f(n -1) with f = 1 -ln(x)2
32·ddx(1 -ln(x)2)·(1 -ln(x)2)(32 -1)
ℹ32·ddx(1 -ln(x)2)·(1 -ln(x)2)(32 -1) = 3·ddx(1 -ln(x)2)·(1 -ln(x)2)(32 -1)2‖ddx(1 -ln(x)2) = 0 + ddx(-ln(x)2)‖32 -1 = 3 -1·22
3·(0 + ddx(-ln(x)2))·(1 -ln(x)2)(3 -1·22)2
We differentiate using: ddx(fn) = n·ddx(f)·f(n -1) (where f = ln(x))
3·(-2·ddx(ln(x))·ln(x))·(1 -ln(x)2)(3 -22)2
ℹ3·(-2) = -6‖ddx(ln(x)) = 1x‖3 -2 = 1
-6·1x·ln(x)·1 -ln(x)22
-6·ln(x)·1 -ln(x)2x2
-6·ln(x)·1 -ln(x)2x2
-6·ln(x)·1 -ln(x)2x2
-6·ln(x)·1 -ln(x)2x·12
-6·ln(x)·1 -ln(x)2x·2
-3·ln(x)·1 -ln(x)2x
You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:
1-3·(1 -ln(x)2)32
-13·(1 -ln(x)2)32
-(1 -ln(x)2)323
Indefinite integrals are defined up to an additive constant C, so this yields: -(1 -ln(x)2)323 + C
▶Antiderivative:   -(-ln(x)2 + 1)323 + C, where C is a constant