Question 356418
{{{ int ( sqrt( (1+x)/(1-x)), dx ) }}}
The problem as you posted it is an indefinite integral. So we will not be evaluating (coming up with a numberic value for) it.<br>
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:
{{{ int ( sqrt( ((1+x)/(1-x))((1+x)/(1+x))), dx ) }}}
This simplifies as follows:
{{{ int ( sqrt( (1+x)^2/(1-x^2) ), dx ) }}}
Using a property of radicals, {{{root(a, p/q) = root(a, p)/root(a, q)}}}, we can split this square root of a fraction into a fraction of square roots:
{{{ int ( sqrt( (1+x)^2)/sqrt(1-x^2) , dx ) }}}
And the numerator simplifies:
{{{ int ( (1+x)/sqrt(1-x^2) , dx ) }}}
(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:
{{{ int ( 1/sqrt(1-x^2)  + x/sqrt(1-x^2), dx )}}}
Now we can use a property of integrals to split this into two:
{{{ int ( 1/sqrt(1-x^2), dx)  + int ( x/sqrt(1-x^2), dx )}}}
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 {{{int (1/sqrt(a^2 - u^2), du)}}} with u = x and a = 1. And the second one is of the form of {{{int (u/sqrt(a^2-u^2), du)}}} 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:
<pre>
   -1
sin  {{{(x/1) + C[1] + (-sqrt(1^2 - x^2)) + C[2]}}}</pre>
Simplifying and combining the two constants of integration into one we get:
<pre>
   -1
sin  {{{(x) + (-sqrt(1 - x^2)) + C}}}</pre>