@Primalory 오래 전 i don't understand why you don't put the +C before going back to the x world, that makes the equality wrong 2026-08-21 19:46 [ytcid:Ugxf2g16NxXPhnuAt_h4AaABAg] i don't understand why you don't put the +C before going back to the x world, that makes the equality wrong
@Bayerwaldler 오래 전 I don’t see that it’s easier. Let u=x-1. Then x(x-1)^1/2 = (u+1)u^1/2 = u^3/2 + u^1/2, dx=du. It’s straightforward from that point on. 2026-08-21 19:14 [ytcid:Ugw87qnfMA4PJ2bHrKl4AaABAg] I don’t see that it’s easier. Let u=x-1. Then x(x-1)^1/2 = (u+1)u^1/2 = u^3/2 + u^1/2, dx=du. It’s straightforward from that point on.
@MisterJackTheA… 오래 전 One day he'll forget the +C, even for just a second. 2026-08-21 18:20 [ytcid:UgzvfWUyOMlOGweyZeJ4AaABAg] One day he'll forget the +C, even for just a second.
@kamalajaiho567… 오래 전 This question just like NCERT textbooks question i am feel that I were solve this type questions in my calss 12 time 2026-08-21 18:15 [ytcid:UgzD4WgriOV-vZIkYdB4AaABAg] This question just like NCERT textbooks question i am feel that I were solve this type questions in my calss 12 time
@STARIslamicPar… 오래 전 Nice explanation 2026-08-21 17:22 [ytcid:Ugz18sFcVJ7ADpEaOyN4AaABAg] Nice explanation
@TahaHashir-o5g 오래 전 The Council invites you 2026-08-22 06:38 [ytcid:UgyZX9H5dIu10IAYqIZ4AaABAg] The Council invites you
@matthewfeig562… 오래 전 Pretty easy to just set u=x-1. The integral becomes Int [(u+1) sqrt u] du = Int [u^(3/2) + u^(1/2)] du. 2026-08-22 04:28 [ytcid:UgxAqb4lWMkoLOBEUit4AaABAg] Pretty easy to just set u=x-1. The integral becomes Int [(u+1) sqrt u] du = Int [u^(3/2) + u^(1/2)] du.
@Necroplantser 오래 전 i tried not using substitution or IBPx√(x-1) = (x-1)√(x-1) + √(x-1)= (x-1)^1.5 + (x-1)^0.5and the rest is history :D 2026-08-22 01:44 [ytcid:UgxhgujoQNqb_wEQa3t4AaABAg] i tried not using substitution or IBP x√(x-1) = (x-1)√(x-1) + √(x-1) = (x-1)^1.5 + (x-1)^0.5 and the rest is history :D
@russelltaylor7… 오래 전 x√(x-1)dxu=√(x-1)u²=x-1x=u²+1dx=2udux√(x-1)dx is replaced by (u²+1)u×2udu=2u²(u²+1)=2u⁴+2u²=2u⁵/5 +2u³/3 + C(2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C(x-1)^3/2 + (x-1)^1/2=(x-1)½[x-1+1]=x√(x-1) ✔️ so x√(x-1)dx = (2/5)(x-1)^(5/2) +(2/3)(x-1)^(3/2) + C 2026-08-22 00:44 [ytcid:UgzurpN9YppjBGPQfTZ4AaABAg] x√(x-1)dx u=√(x-1) u²=x-1 x=u²+1 dx=2udu x√(x-1)dx is replaced by (u²+1)u×2udu= 2u²(u²+1)=2u⁴+2u²=2u⁵/5 +2u³/3 + C (2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C (x-1)^3/2 + (x-1)^1/2=(x-1)½[x-1+1]=x√(x-1) ✔️ so x√(x-1)dx = (2/5)(x-1)^(5/2) +(2/3)(x-1)^(3/2) + C
@julianheller27… 오래 전 Dieses Integral könnte man auch mit der partiellen Integration lösen. (D: x ;I: √(x-1)). Aber etwas unangenehm mit den neuen Exponenten 3/2 und 5/2. 2026-08-22 00:42 [ytcid:Ugxqo8URLkJttyPjp7d4AaABAg] Dieses Integral könnte man auch mit der partiellen Integration lösen. (D: x ; I: √(x-1)). Aber etwas unangenehm mit den neuen Exponenten 3/2 und 5/2.
@brainfellow514… 오래 전 That was actually easier than it looked. Nicely done. 2026-08-21 22:34 [ytcid:UgyIxioqJwMmoe_4q8Z4AaABAg] That was actually easier than it looked. Nicely done.
@davidmilhousca… 오래 전 What happens to the exponents? Why/how does it work? 2026-08-22 11:02 [ytcid:UgwiCtxys2n1yPUGbBl4AaABAg] What happens to the exponents? Why/how does it work?
@Mr.Chickem 오래 전 Do I have to memorize all the series tests 2026-08-22 10:12 [ytcid:UgyQ1IWnF9DNx87Ubjd4AaABAg] Do I have to memorize all the series tests
@Tetraverse 오래 전 Please make content on probability theory for one of these 2026-08-22 09:36 [ytcid:UgzW07qGGHtBKGlsjB14AaABAg] Please make content on probability theory for one of these