Evaluet the integral?

Question

#int_1^9 (4+x^2)/sqrt(x) dx#

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Answer 1 · 2018-04-19T23:32:11

#int_1^9 (4+x^2)/sqrtx dx = 564/5 #

Substitute:

#t= sqrtx#

#dt = dx/(2sqrtx)#

so:

#int_1^9 (4+x^2)/sqrtx dx = 2 int_1^3 (4+t^4)dt#

using the linearity of the integral:

#int_1^9 (4+x^2)/sqrtx dx = 8int_1^3 dt + 2 int_1^3 t^4dt#

#int_1^9 (4+x^2)/sqrtx dx = 8[t]_1^3 dt + 2 [t^5/5]_1^3 #

#int_1^9 (4+x^2)/sqrtx dx = 24-8 + 2/5 (243-1) #

#int_1^9 (4+x^2)/sqrtx dx = 564/5 #