TITLE: How do I solve $\int\frac{xd(1-x^2)}{1-x^2}$?
QUESTION [0 upvotes]: It might be an easy question, but I want to know how to solve an integration problem when the differential in the integration (sorry don't know its actual name) is not $\int d(x)$, like in the question above what if it's
$$\int d(x^2)$$
or
$$\int d(x^3 - 1)$$
Also, does this way of solving problems have a name or something?
Thanks in advance.

REPLY [0 votes]: See the $$\int x\frac{d\boxed{1-x^2}}{\boxed{1-x^2}}$$
Since $$\frac d{dx} (1-x^2)=\frac{d(1-x^2)}{dx}=-2x\implies d(1-x^2)=-2x dx$$
Therefore $$\int x\frac{d(1-x^2)}{1-x^2}= \int x\frac{-2x dx}{1-x^2} =2\int \frac{x^2}{x^2-1}dx$$
You can also try integration by parts:
$$\int u\,dv+\int v\,du=uv\mathop \implies^{ v=1-x^2, u=\frac{x}{1-x^2}}\int \frac{x}{1-x^2}d(1-x^2)= \frac{x}{1-x^2} (1-x^2)-\int(1-x^2)d\left(\frac{x}{1-x^2}\right)$$
Now use the Fundamental theorem of Calculus and use the differential:
$$x-\int d\left(\frac{x}{1-x^2} \right)-\int x^2d\left(\frac{x}{1-x^2}\right)=x+\frac{x}{x^2-1}-\int x^2d\left(\frac{x}{1-x^2}\right)= x+\frac{x}{x^2-1}-\int x^2\frac{x^2+1}{(x^2-1)^2}dx= x+\frac{x}{x^2-1}-\int \frac{x^4}{(x^2-1)^2}dx-\int \frac{x^2}{(x^2-1)^2}$$
Can you do one of the integrals from here?
Please correct me and give me feedback!
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