Question #6d541

Question

903 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-15T17:24:12

#e^x(x^2+1)+c#

Take the integral

#inte^x(x+1)^2dx#

Expanding the integrand #(x+1)^2=x^2+2x+1#

#int(e^x x^2+2e^x x+e^x)dx#

separate

#inte^x x^2dx+int2e^x xdx+inte^xdx#

For the integrand #e^x x^2#, integrate by parts,
#intfdg=fg-intgdf#

#f=x^2, dg=e^xdx#
#df=2xdx# #g=e^x#

#e^x x^2-inte^x 2xdx#

The integral of #e^x=e^x+c#

#inte^x x^2dx+int2e^x xdx+inte^xdx=e^x x^2-inte^x 2xdx+int2e^x xdx+e^x+c#

#inte^x x^2dx+int2e^x xdx+inte^xdx=e^x x^2+e^x+c#

#inte^x x^2dx+int2e^x xdx+inte^xdx=e^x(x^2+1)+c#