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

1 Answer

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

Explanation:

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