How do you integrate #int 1/(sqrtx(1+sqrtx)^2dx# from [1,9]?

1 Answer
Jan 14, 2017

#int_1^9 1/(sqrtx(1+sqrtx)^2)dx=1/2#

Explanation:

#I=int_1^9 1/(sqrtx(1+sqrtx)^2)dx#

We will use the substitution #u=1+sqrtx#. Differentiating this reveals that #du=1/(2sqrtx)dx#. We can rewrite the integral to make both of these evident:

#I=2int_1^9(1+sqrtx)^(-2)(1/(2sqrtx)dx)#

Before switching from #dx# to #du#, change the variables of #x=1# and #x=9# by plugging them into #u=1+sqrtx#.

#x=1# becomes #u=1+sqrt1=2# and #x=9# becomes #u=1+sqrt9=4#. Then:

#I=2int_2^4u^-2du#

Using #intu^ndu=u^(n+1)/(n+1)#:

#I=2[u^(-1)/(-1)]_2^4=2[-1/u]_2^4=2(-1/4-(-1/2))#

#I=2(-1/4+1/2)=2(1/4)=1/2#