What is #f(x) = int 1/(sqrt(x+3) dx# if #f(1)=7 #?

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Answer 1 · 2016-01-19T09:27:58

#f(x)=2 sqrt(x+3) + 3#

I believe that the given must be like
#f(x)= int (1/sqrt(x+3)) dx# and #f(1)=7#

Also #f(x)=int (x+3)^(-1/2)*dx#
The integral becomes like this after integration

#f(x)=((x+3)^(-1/2+1))/(-1/2+1)+C#

which simplifies to

#f(x)=((x+3)^(1/2))/(1/2)+C#

and

#f(x)=2 sqrt(x+3)+C#

but #f(1)=7#

#f(1)=7=2 sqrt(1+3)+C#

#7=2*sqrt(4)+C#
#7=2*2+C#
#C=3#

Finally

#f(x)=2 sqrt(x+3)+3#