How do you integrate #int(e^x + 1)^2dx#?

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Answer 1 · 2017-06-04T14:35:13

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!

Answer 2 · 2017-06-04T14:46:41

#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)