How do you evaluate the definite integral #int((sqrtx + 1)/(4sqrtx))^2 dx # from #[3,9]#?

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Answer 1 · 2016-07-06T10:23:45

#int_3^9 ((sqrtx+1)/(4sqrtx))^2*dx=#
#9/8-sqrt3/4+1/16*ln 3=0.7606505661495#

From the given, #int_3^9 ((sqrtx+1)/(4sqrtx))^2*dx#
We start by simplifying first the integrand

#int_3^9 ((sqrtx+1)/(4sqrtx))^2*dx#

#int_3^9 ((sqrtx)/(4sqrtx)+1/(4sqrtx))^2*dx#

#int_3^9 (1/4+1/(4sqrtx))^2*dx#

#int_3^9 (1/4)^2*(1+1/(sqrtx))^2*dx#

#int_3^9 (1/16)*(1+2/(sqrtx)+1/x)dx#

#(1/16)*int_3^9 (1+2*x^(-1/2)+1/x)dx#

#(1/16)* [x+(2*x^(1/2))/(1/2)+ln x]_3^9#

#(1/16)* [x+4*x^(1/2)+ln x]_3^9#

#(1/16)* [(9+4*9^(1/2)+ln 9)-(3+4*3^(1/2)+ln 3)]#

#(1/16)*[9+12+ln 9-3-4sqrt3-ln 3]#

#(1/16)(18-4sqrt3+ln 3)#

#9/8-sqrt3/4+1/16*ln 3#

#0.7606505661495#

God bless....I hope the explanation is useful.