First, find the indefinite integral and then evaluate from the lower bound to the upper bound. Explanation: \displaystyle{\int_{{0}}^{\sqrt{{{7}}}}}\frac{{{2}{x}}}{{\left({1}+{x}^{{2}}\right)}^{{\frac{{1}}{{3}}}}}{\left.{d}{x}\right.}=? ...

\displaystyle{\ln}{\left|{x}-{1}\right|}+{2}{\ln}{\left|{x}+{4}\right|}+{C} , or, \displaystyle{\ln}{\left|{\left({x}-{1}\right)}{\left({x}+{4}\right)}^{{2}}\right|}+{C} . Explanation: Let \displaystyle{I}=\int\frac{{{3}{x}+{2}}}{{{x}^{{2}}+{3}{x}-{4}}}{\left.{d}{x}\right.}=\int\frac{{{3}{x}+{2}}}{{{\left({x}-{1}\right)}{\left({x}+{4}\right)}}}{\left.{d}{x}\right.} ...

You actually didn't use the chain rule correctly: To use the chain rule, you need to also multiply 2(1 - x) by \frac d{dx}(1 - x)= -1. In other words, we have f(x) = (1 - x)^2 \implies f'(x) = 2(1 - x)\cdot \frac{d}{dx}(1 - x) = -2(1 - x) = 2(x - 1)

The computer science community has a vast literature on this topic because when you are implementing systems you don't have the option of being lax about what exactly you mean by a binding form. For ...

You can differentiate the second expression "directly" because of the power rule for monomials and because the derivative operator is linear: the derivative of a sum is the sum of the derivatives, and ...

Let us generalise a bit. The integral of (a+bx^n)^m is given by (just expanding and integrating term-wise): \int(a+bx^n)^m\,\mathrm dx = \sum_{k=0}^m \binom m ka^{m-k}b^k\frac{x^{kn+1}}{kn+1} ...