Question

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

Answer 1

`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`