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

1 Answer
Mar 20, 2016

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