How do you evaluate the integral #(4+sqrt(x))^9 dx# using u substitution?

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Answer 1 · 2015-04-20T10:22:40

Hi !
#1^(st)# method : You can use #(a+b)^n=sum_(k=0)^(n=9)(""_k^n)a^(n-k)b^k#
And integrate directly

#2^(nd)# method :

#u = 4+sqrt(x)#

#(du)/dx = 1/(2sqrt(x)#

#1/dx=1/(2sqrt(x))*1/(du)#

#dx = 2sqrt(x) du = 2u-8du#

#=>int(2u-8)u^9du#

#=>int2u^10-8u^9du#

#[2/11u^11-8/10u^10]#

#[2/11(4+sqrt(x))^11-8/10(4+sqrt(x))^10]+C#