Is there a smarter way to integrate? My textbook uses the following technique to integrate. Let us take the following as an example,  evaluate the integral of:
$$f(x) = 6x(x^2 + 1)^5$$
Notice that $[x^2+1]' = 2x$, so $6x\space dx = 3 \space d (x^2+1)$
$$6x(x^2+1)^5 \space dx = \dfrac{6}{2} (x^2+1)^5 \space d(x^2+1) = 3u^5 \space du = d \dfrac{1}{2} u^6 = d \dfrac{1}{2} (x^2+1)^6$$
So $$F(x) = \dfrac{1}{2} (x^2 +1)^6 + c$$
This method seems very confusing (what is the meaning of the $d$ for example?) and besides, it seems like it would work only in textbook-cases. Is there a 'better' way of integrating these types of basic functions?
p.s. - I looked in my textbook for reference, but the toughest types of functions we have to evaluate are the integrals of for example $h(x) = \dfrac{2x-3}{x^2+6x+10}$ and $k(x) = \dfrac{\arcsin^2(x)}{\sqrt{1-x^2}}$. So the method wouldn't of course have to work for integrals tougher than these types. 
 A: First of all, the following boils down to exactly the same as your textbook does, but it might make more sense to you. Make the substitution $u = x^2 +1$. Then $du= 2xdx$. Hence $\int 6x(x^2+1)^5dx=\int 3 u^5du$. Now integrate and substitute back for $u$.
Notice that it was a very plausible idea to make this substitution because we see the the integral involves a product of a certain function ($u$) multiplied by the derivative of this function ($du$).
A: Precise meanings of these statements are not usually explained in calculus classes from what I've seen.
It makes more sense when you learn some differential geometry: the $d$ is just the exterior derivative of a differential ($0$-)form.
More directly, it can be seen as just a shorthand notation for integration by substitution, which usually is explained in calculus classes. And frequently, it is the most convenient way of explicit integration (well, at least for things you're supposed to integrate during calculus courses, anyway).
A: There are many dedicated methods to solve different integration problem.The method you have shown here is fundamental method which is not directly used in complex problem but it will use in many complex problem when that problem reduces to easier form. d comes from differentiation.This is substitution example.we differentiate it to reduce it easier form
