∫ xex(1+x)2dx=∫ (x+1−1)ex(1+x)2dx=∫ ex1+xdx−∫ ex(1+x)2dx=\int \:\frac{xe^x}{\left(1+x\right)^2}dx =\int \:\frac{(x+1-1)e^x}{(1+x)^2}dx =\int \:\frac{e^x}{1+x}dx -\int \:\frac{e^x}{(1+x)^2}dx =∫(1+x)2xexdx=∫(1+x)2(x+1−1)exdx=∫1+xexdx−∫(1+x)2exdx=
∣u=1x+1,dv=exdx,du=−1(1+x)2dx,v=ex∣|u=\frac{1}{x+1}, dv=e^xdx, du=-\frac{1}{(1+x)^2}dx, v=e^x|∣u=x+11,dv=exdx,du=−(1+x)21dx,v=ex∣
=11+xex+∫ ex(1+x)2dx−∫ ex(1+x)2dx=11+xex+C=\frac{1}{1+x}e^x+\int \:\frac{e^x}{(1+x)^2}dx -\int \:\frac{e^x}{(1+x)^2}dx =\frac{1}{1+x}e^x+C=1+x1ex+∫(1+x)2exdx−∫(1+x)2exdx=1+x1ex+C
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments