How do you integrate #int(e^x + 1)^2dx#?
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The integral equals # 1/2e^(2x) + 2e^x + x + C#
Expand:
#I = int(e^(2x) + 2e^x + 1)dx#
Separate the integrals:
#I = inte^(2x)dx + 2inte^xdx + int 1 dx#
Use #int(e^x) = e^x#:
#I = 1/2e^(2x) + 2e^x + x + C#
Hopefully this helps!
#1/2e^(2x)+2e^x + x + "constant"#
This is a good opportunity to use #u#-substitution. You just might have to use it twice though.
Let #u=e^x# and #du=e^xdx => 1/e^xdu=dx => 1/udu=dx#
So #int(e^x+1)^2dx# becomes #int(u+1)^2/(u)du#
Let #s=u+1# so that #ds=du#
So #int(u+1)^2/(u)du# becomes #ints^2/(s-1)ds#
By long division,
#ints^2/(s-1)ds=int(s+1/(s-1)+1)ds#
Integrate each separately,
#=intsds+int1/(s-1)ds+intds#
#=s^2/2+ln(s-1)+s+"constant"#
Plugging in #s=u+1# gives
#=(u+1)^2/2+ln(u)+u+1+"constant"#
Plugging in #u=e^x# gives
#=(e^x+1)^2/2 + ln(e^x) + e^x + 1 + "constant"#
#=(e^(2x)+2e^x+1)/2 + x + e^x + 1 + "constant"#
#=e^(2x)/2+e^x+1/2 + x + e^x + 1 + "constant"#
#=1/2e^(2x)+2e^x + x + 3/2 + "constant"#
#=1/2e^(2x)+2e^x + x + "constant"# (where #3/2# is absorbed into the constant)
