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

1 Answer

Jan 30, 2017

#1/15(2x+1)^(3/2)(3x-1)+C.#

Explanation:

Let #2x+1=t^2 :. x=(t^2-1)/2, and, dx=tdt.#

Therefore, the Reqd. Integral#=int{((t^2-1)/2)(sqrtt^2)(t)}dt#

#=1/2int(t^4-t^2)dt=1/2(t^5/5-t^3/3)=t^3/30(3t^2-5)#

#=1/30(t^2)^(3/2)(3t^2-5)#

#=1/30(2x+1)^(3/2){3(2x+1)-5}#

#=1/15(2x+1)^(3/2)(3x-1)+C.#

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