Evaluating an integrals by appropriate substitution i can't understand how to solve this issue: using an appropriate substitution, evaluate this integral:
$$
\int \frac{1+x²}{\sqrt{1+x}}\mathrm{d}x
$$
can any one solve this so i can understand how to do this.
 A: When you are facing a radical which has a linear sum in it, it works well to use that as the basis of the substitution.  Here, you would take $ \ u = x + 1 \ $ , which will give you $ \ du = dx \ $ .  To deal with the numerator, you need to solve your substitution equation for $ \ x \ $ , giving $ \ x = u - 1 \ $ .  The integral becomes
$$ \int \  \frac{1+ x^2}{\sqrt{1 + x }} \ dx \ \rightarrow \ \int \frac{1 + (u - 1)^2}{\sqrt{u}} \ du  \ . $$ 
You would then multiply out the polynomial in the numerator.
The point in doing this is that you now have a polynomial divided simply by the square root of the variable, which will leave you with a set of terms in the integrand which are all just fractional powers of $ \  u \ $ , something which is much easier to integrate.
A: Let $u^2=1+x$. Then $2u\,du =dx$. Using the fact that $x=u^2-1$, we find that $x^2+1=u^4-2u^2+2$. 
Substituting, we find that our integral is
$$\int \frac{(u^4-2u^2+2)(2u)}{u}\,du.$$
There is cancellation, and we have reached
$$\int (2u^4-4u^2+4)\,du$$
which is
$$\frac{2}{5}u^5 -\frac{4}{3}u^3+4u+C.$$
Now replace $u$ by $(x+1)^{1/2}$.
