Simple Integration by Substitution I am just exercising integration by substitution and would like to solve the following integral:
$$\int (x^2+1)^3 dx$$
I substitute $x^2$ with y and hence $\frac{dy}{dx} = 2x$ and $dx = dy*2x$. Thus I get:
$$\int (y+1)^3 dy*2x$$
how do I continue from this point on, i.e. how do I get rid of the 2x?
 A: Here, the substitution isn't appropriate, because, as you see, there's no factor of $x$ in the original integrand.  It would have been a better choice to substitute $y = (x^2 + 1)$ if the integral had been $$\int x(x^2 + 1)^3\,dx$$ because then we'd have $\;dy = 2x dx\implies dx = \frac 12 dy,\;$ giving us a very nice integral to work with: $$\;1/2 \int y^3 dy$$
But, alas! We don't have that integral to work with. And there's not a really handy substitution to use that will simplify our work. 
Instead, for this integral, try expanding the binomial (easy to do in this case), and use the power rule to integrate each term:
$$\int (x^2+1)^3 dx \quad = \quad \int (x^6 + 3x^4 + 3x^2 + 1) \,dx \quad = \quad\dfrac{x^7}{7} + \frac{3x^5}{5} + x^3 + x + C$$
A: $\frac{dy}{dx} = 2x$ implies $dx = \frac{dy}{2x}$.
To get rid of the $x$, recognize that $x^2 = y$ implies $x = \pm \sqrt{y}$.
A: You should not perform substitution here because the derivative $2x$ was not there to begin with, making for a difficult process. I suggest actually expanding out $(x^2+1)^3$ into a polynomial, which then is very easy to integrate.
