how do we integrate [ (x e^x)/ ( 1+x )^2

Asked By emma On 03/05/2017 09:34

Answers

Hi Emma! I will try to walk through this problem for you step by step. We will need to use integration by parts for this problem and u substitution. Integration by parts: ?(u*dv) = v*u - ?(v*du) so our goal is to find each of these parts where u=xe^x, dv = (x+1)^-2dx (see below) and we will find v and du 1. Rewrite the problem as multiplication instead of division: ?( (xe^x)*(x+1)^-2)dx 2. let u = xe^x du = x(e^x) + e^x(1)dx = e^x(1+x)dx 3. dv = (x+1)^-2dx so integrate this now, again using substitution to find v ?((x+1)^-2)dx let u = x+1 so then du = (1)dx so substituting, we get; ?( u^-2)du = u^(-2+1)/(-2+1) = -u^-1 + c = -(x+1)^-1 +c So v= -(x+1)^-1 + c 4. Now plug in the pieces to the integration by parts u = 1-x dv = (x+1)^-2dx v = -(x+10^-1 = -1/(x+1) du = e^x(1+x)dx ?(udv) = vu - ?(vdu) ?(xe^x*(x+1)^-2)dx = -1/(x+1)*xe^x - ?((-1/x+1)*e^x(1+x)dx) now you can cancel out the x+1 and 1+x in the integral to get: =( -xe^x)/(x+1) - ? (-e^x)dx =(-xe^x)/(x+1) + ? (e^x)dx =(-xe^x)/(x+1) + e^x + c now combine fractions by finding like denominators = (-xe^x + e^x(x+1))/(x+1) +c distribute the e^x =(-xe^x+xe^x+e^x)/(x+1)+c factor out e^x from the numerator =e^x(-x +x +1)/(x+1) + c =e^x(1)/(x+1) + c = e^x/(x+1) + c So, ? [x e^x /(x + 1)²] dx = [e^x /(x + 1)] + c I hope this helps! : ) Mallory
Answered On 03/05/2017 23:26