Can this be solved? #int ((sqrtx+1)/(2sqrtx))^4dx#

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P.S. cause my teacher give us practice paper that include this question but the answer of it is solving for the integral #int (sqrtx+1)^4/(2sqrtx)dx# instead. I would like to learn more and see if the integral I ask can be solved or not. If it can be solved, I would like to know the steps. Thank you :)

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Answer 1 · 2017-12-04T10:34:02

# 1/16(-1/x-8/sqrtx+6ln|x|+8sqrtx+x)+C.#

#I=int{(sqrtx+1)/(2sqrtx)}^4dx,#

#=1/2^4int{(sqrtx+1)/sqrtx}^4dx,#

#=1/16int(sqrtx/sqrtx+1/sqrtx)^4dx,#

#=1/16int(1+x^(-1/2))^4dx,#

#=1/16int{(x^(-1/2))^4+4(x^(-1/2))^3+6(x^(-1/2))^2+4(x^(-1/2))^1+1}dx,#

#=1/16int{x^(-2)+4x^(-3/2)+6x^(-1)+4x^(-1/2)+1}dx,#

Now, #intt^ndt=t^(n+1)/(n+1), n ne-1, &, intt^(-1)dt=ln|t|.#

#:. I=1/16(x^(-1)/-1+4*x^(-1/2)/(-1/2)+6ln|x|+4*x^(1/2)/(1/2)+x),#

#rArr I=1/16(-1/x-8/sqrtx+6ln|x|+8sqrtx+x)+C.#