Integral of x dx / sqrt(x+2)
Make the substitution sqrt(x+2)=u
(x+2)^(1/2) = u
(1/2)(x+2)^(-1/2) dx = du
1/2(x+2)^(1/2) dx = du
1/2sqrt(x+2) dx = du
1/sqrt(x+2) dx = 2 du
Integral of x dx / sqrt(x+2) = Integral 2 x du
sqrt(x+2) = u
(x+2)=u^2
x=u^2-2
Integral 2 x du = Integral 2(u^2-2) du
= Integral 2u^2 du - 4 du
= 2 u^3/3 - 4u + C
= (2/3) (x+2)^(3/2) - 4 sqrt(x+2) + C
[The sqrt(3)+sqrt(2)]/sqrt(6)can be solved by multiply the numerator and denominator by sqrt(6)This gives sqrt(18)+sqrt(12)/6This can be simplified to[3(sqrt(2)+2(sqrt(3)]/6
(5 + sqrt(5)) / 2 = 3.61803399
3
4
x*sqrt(2)/{5 - sqrt(3)} = {5 + sqrt(3)} => x*sqrt(2) = {5 + sqrt(3)} * {5 - sqrt(3)} = 25 - 3 = 22 => x = 22/sqrt(2) = 22*sqrt(2)/{sqrt(2)*sqrt(2)} = 22*sqrt(2)/2 = 11*sqrt(2)
[The sqrt(3)+sqrt(2)]/sqrt(6)can be solved by multiply the numerator and denominator by sqrt(6)This gives sqrt(18)+sqrt(12)/6This can be simplified to[3(sqrt(2)+2(sqrt(3)]/6
(5 + sqrt(5)) / 2 = 3.61803399
3
4
x*sqrt(2)/{5 - sqrt(3)} = {5 + sqrt(3)} => x*sqrt(2) = {5 + sqrt(3)} * {5 - sqrt(3)} = 25 - 3 = 22 => x = 22/sqrt(2) = 22*sqrt(2)/{sqrt(2)*sqrt(2)} = 22*sqrt(2)/2 = 11*sqrt(2)
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
No
Vrms=sqrt[1/T * integral(v^2(t)dt, 0,t] Irms=sqrt[1/T * integral(i^2(t)dt, 0,t]
sqrt(8)/[2*sqrt(3)] = 2*sqrt(2)/[2*sqrt(3)] = sqrt(2)/sqrt(3) = sqrt(2/3)
Approximately 2.828427
(7*sqrt(2))/2 (Seven time the square root of 2, divided by two)
0.2857