Question

What is `f(x) = int x/sqrt(x^2+1) dx` if `f(2) = 3`?

Answer 1

`f(x)=sqrt(x^2+1)+3-sqrt5`

Explanation:

To integrate:

Let `u=x^2+1` so `du=2xdx`.

This gives us:

`f(x)=1/2int(2x)/sqrt(x^2+1)dx=1/2int1/sqrtudu=1/2intu^(-1/2)du`

Now, we integrate using the rule: `intu^ndu=u^(n+1)/(n+1)`

So, we have

`f(x)=1/2u^(1/2)/(1/2)+C=1/2sqrtu*2+C=sqrtu+C`

`f(x)=sqrt(x^2+1)+C`

Using the original condition `f(2)=3`, we have

`3=sqrt(2^2+1)+C`

`3=sqrt5+C`

`C=3-sqrt5`

So, substituting this in, we see that

`f(x)=sqrt(x^2+1)+3-sqrt5`