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You need to use integration by parts so that (cap. S stands for ingegral)
Su*dv = uv - Sv*du
So for Sxln(x) you need to choose your u and dv, then use those to solve for du and v. ln(x) is not easy to integrate, but it is easy to differentiate, so choose it as your "u", therefore xdx would be "dv"
u = ln(x); dv = xdx
du = (1/x)dx; v = 1/2*x^2
So now you just plug it into the formula above
Sudv = uv - Svdu
= ln(x)*(x^2/2) - S(x^2/2)*(dx/x) --- pull out the 1/2, and simplify x^2/x
= ln(x)*(x^2/2) - 1/2*[Sxdx]
= 1/2*(x^2ln(x)) - 1/2*[1/2*x^2]
= 1/2*(x^2ln(x)) - 1/4(x^2) + C
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What is the Derivative of x to the power of x?
The solution to this is: (xx)'= (elnx to the power of x)'= (exlnx)'= (xlnx)'*exlnx= [x(1/x) + 1(lnx)]*exlnx = (lnx+1)*exlnx= (lnx+1)*xx
Simplify e to the 1 plus Inx power?
e1 + (lnx) = e1 * e(lnx) = e * x = ex
What is the derivative of lnx squared?
I do not see why the chain rule would not work here. d/dx (inx)^2 = 2(lnx) * 1/x = 2(lnx)/x
What is the integral of x.lnx?
Do you want ∫x lnx dx? Let's call that I, which we now seek to find. The solution is I = ½ x2 lnx - ¼ x2; here is how we can find it: Let z = lnx. Then, x = ez, dx = ez dz, and dI = x lnx dx. Then, dI = (ez)(z)(ez dz) = ze2z dz = ¼ (2z e2z d(2z)). Thus, 4dI = wew dw, where we let w = 2z = 2 lnx. Now, d(wew) = ew dw + w d(ew) = ew dw + wew dw = ew dw + 4dI; hence, 4dI = d(wew) - ewdw = d(wew) - d(ew)­ = d((w - 1)ew). Therefore, 4I = (w - 1)ew = (2 lnx - 1)x2 = 2x2 lnx - x2; and I = ½ x2 lnx - ¼ x2, which is the answer we sought. Checking, we differentiate back, to confirm that I' = x lnx: I = ¼ x2(2 lnx - 1), whence, 4I' = (x2(2 lnx - 1))' = 2x(2 lnx - 1) + x2 (2/x) = 2x(2 lnx - 1) + 2x = 2x(2 lnx) = 4x lnx; thus, I' = x lnx, re-assuring us that we have integrated correctly.
What is the derivative of 1 lnx?
The derivative of 1/lnx, can be found easily using either the chain rule or the quotient rule. It is -1/[x*(lnx)2]
