How do you integrate #int sqrt(1+2x)# using substitution?

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How do you integrate #int sqrt(1+2x)dx# using substitution?

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Answer 1 · 2016-09-09T21:10:41

#1/3(sqrt(1+2x))^3+C, or#

#1/3(1+2x)^(3/2)+C#.

We subst. #1+2x=t^2" in "I=intsqrt(1+2x)dx#.

#:. 2dx=2tdt, or, dx=tdt#. Also, #sqrt(1+2x)=t#

Therefore, #I=intt*tdt=intt^2dt=t^3/3#

#=1/3(sqrt(1+2x))^3+C, or#

#intI=1/3(1+2x)^(3/2)+C#.