integrate function with change of variable Find the primitive of $\;\displaystyle \int x^2 \sqrt{x+1}$ $dx$
So (...)
$u = x + 1 \quad \iff \quad u - 1 = x$
$u' = 1 \quad \iff \quad \frac{du}{dx} = 1 \rightarrow \;du = dx$
$$\int(u - 1)^2 . u^\frac12 \; du \;\;= \;\; {{(u-1)^3}\over3} \cdot {u^{3/2}\over{3/2}} + C \;\; =\;\; {1\over3} x^3\cdot{2\over3}(x+1)^{3/2} + C$$ 
Is this correct?
 A: The substitutions will work fine...but your evaluation is problematic. 
Most problematic is the fact that you are integrating each factor of a product in the integrand, and expressing this as the product of integrated factors: which you cannot do, unless you are using, say, integration by parts, which proceeds much differently: . I.e. $$\int [f(x)\cdot g(x)]\,dx \;\neq \;\int f(x) \,dx \cdot \int g(x)\,dx$$

So, let's start from the point after which we've substituted:
$$\int \underbrace{(u - 1)^2}_{\text{expand}} . u^{1/2} \; du \; = \;\int \underbrace{(u^2 - 2u + 1)u^{1/2}}_{\text{distribute}} \,\;du = \int \left(u^{5/2} - 2u^{3/2} + u^{1/2}\right) \,du$$
Now integrate, and then back-substitute.
A: You have it right up to $$\int(u-1)^2 u^{1/2} du$$. However it is not in general correct that $\int (f\cdot g)(x)dx=\int f(x)dx \cdot \int g(x) dx $, that is, integration does not "distribute" over multiplication.
A: No, it isn't. If you take the derivative of the product applying Leibniz's rule it won't give you the original integrand. The error is that you integrated both parts of the functions instead of the function as a whole.
The integral is easier if you make the change of variable: 
$$u=\sqrt{1+x}$$
Then:
$$du=\frac{dx}{2\sqrt{1+x}}$$
So the integral becomes:
$$2\int u^2(-1+u^2)^2 du$$
And now it's an inmmediate integral.
