You can put this solution on YOUR website!
The problem as you posted it is an indefinite integral. So we will not be evaluating (coming up with a numberic value for) it.
The "trick" to a lot of integration problems is to use Algebra (and perhaps Trig) to transform the integral you have into one or more integrals which fit a pattern you know how to identify.
With our integral, I am going to start by multiplying the numerator and denominator so that the numerator is a perfect square:
This simplifies as follows:
Using a property of radicals, , we can split this square root of a fraction into a fraction of square roots:
And the numerator simplifies:
(From the original function we know that 1+x must be positive so we do not need to be concerned with +- issues on this square root.)
We can now split this into separate fractions:
Now we can use a property of integrals to split this into two:
By doing this manipulation of the integral we now have two integrals in a form we should be able to solve. The first one is of the form with u = x and a = 1. And the second one is of the form of with u = x and a = 1. (In your Calulus book there is probably a table of common forms. You should be able to find both of these in this table so you can see where the solution came from:
-1
sin
Simplifying and combining the two constants of integration into one we get:
(