5/5x + c where c is the constant of intergration just differentiate the 5x to get 5 and times that by 1/5x then add c The answer above is wrong. This is simply because you cannot intergrate lnx to get 1/x. To intergrate I would recommend using intergration by parts. u=ln5x du/dx=1/x dv/dx=1 v=x uv-(intergal of)v.du/dx =xln5x-intergral of x/x intergral of x/x = x =xln5x-x+c = x(ln5x-1)+c
x
X to the fourth power is... X times X times X times X
By definition of powers, x to the power y = x*x*x*........y times x to the power 1 = x Therefore, 10 to the power 1 = 10.
When multiplying two terms with the same base, you add the exponents. In this case, x to the fifth power times x to the third power would be x to the power of 5 + 3, which simplifies to x to the power of 8. So, x to the fifth power times x to the third power equals x to the eighth power.
Use integration by parts. integral of xe^xdx =xe^x-integral of e^xdx. This is xe^x-e^x +C. Check by differentiating. We get x(e^x)+e^x(1)-e^x, which equals xe^x. That's it!
{xe^x dx integrate by parts let f(x) = x so f'(x) = 1 and g(x) = e^x so g'(x) = e^x so.. f(x)*g(x) - {(g(x)*f'(x)) dx therefore xe^x - {(e^x * 1) dx so.. xe^x - e^x + C factorize so... (x-1)e^x + C
let u = x du=dx let dv= e^x v=e^x ∫ xe^(x)dx = xe^x - ∫ e^(x)dx = xe^x - e^x = e^x ( x-1 ) + c
xe^(5x) is an expression involving x and Euler's constant
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
Xenon (Xe)
Depending on which software you have,you may be able to intergrate Quickbooks.
xe-x
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. 10.0 liters of Xe gas at STP would therefore contain 10.0/22.4 = 0.4464 moles of Xe. 1 mole of Xe contains 6.022 x 10^23 atoms. Therefore, 10.0 liters of Xe gas at STP would contain 0.4464 x 6.022 x 10^23 = 2.69 x 10^23 xenon atoms.
featured, intergrate
There is no element that starts with the letter "X" but there is an "Xe" which is Xenon.Xenotime
5/5x + c where c is the constant of intergration just differentiate the 5x to get 5 and times that by 1/5x then add c The answer above is wrong. This is simply because you cannot intergrate lnx to get 1/x. To intergrate I would recommend using intergration by parts. u=ln5x du/dx=1/x dv/dx=1 v=x uv-(intergal of)v.du/dx =xln5x-intergral of x/x intergral of x/x = x =xln5x-x+c = x(ln5x-1)+c