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

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Answer 1 · 2016-03-20T20:51:36

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

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#