Question 127249
{{{int( 2xe^(x^2), dx, 0, 1 )}}} Start with the given integral


Let {{{u=x^2}}}. So this means that {{{du=2xdx}}}. So the {{{x^2}}} gets replaced with {{{u}}} and the {{{2xdx}}} (you have to group these together) gets replaced with {{{du}}} 


Also, since the first endpoint is 0, this means that {{{x=0}}}. But we want this in terms of u. So just plug in {{{x=0}}} into {{{u=x^2}}} to get {{{u=0}}}. Do the same thing with the other endpoint to get {{{u=1}}}


So we now have:

{{{int( e^u, du, 0, 1 )}}}



So we now have:

{{{e^u+C}}} Take the integral of {{{e^u}}} to get {{{e^u}}}. Remember {{{int( e^u, du)=e^u+C}}}



Now let's evaluate the endpoints



{{{(e^1+C)-(e^0+C)}}} Plug in the endpoints and subtract



{{{e-1}}} Combine like terms and simplify




So {{{int( 2xe^(x^2), dx, 0, 1 )=e-1}}}




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Are you trying to evaluate {{{int( 1/x, dx)}}} ?



The integral of {{{1/x}}} is {{{ln(x)+C}}}. In other words, {{{int( 1/x, dx)=ln(x)+C}}}