\displaystyle{\int_{{2}}^{{4}}}{\left({3}{x}^{{4}}\right)}{\left.{d}{x}\right.}=\frac{{2976}}{{5}} Explanation: \displaystyle{\int_{{2}}^{{4}}}{\left({3}{x}^{{4}}\right)}{\left.{d}{x}\right.} ...

\displaystyle{x}^{{5}}+{C} Explanation: \displaystyle\int{\left({5}{x}^{{4}}\right)}{\left.{d}{x}\right.}={5}\int{\left({x}^{{4}}\right)}{\left.{d}{x}\right.} Applying the power rule: \displaystyle={5}\cdot\frac{{x}^{{5}}}{{5}}+{C}={x}^{{5}}+{C}

\displaystyle\frac{{81}}{{4}} Explanation: \displaystyle{\int_{{0}}^{{3}}}{x}^{{3}}{\left.{d}{x}\right.}=\frac{{x}^{{4}}}{{4}}{\mid}^{{{3},{0}}} \displaystyle=\frac{{3}^{{4}}}{{4}}-{0} ...

The answer is \displaystyle=\frac{{x}^{{6}}}{{6}}+{C} Explanation: We use \displaystyle\int{x}^{{n}}{\left.{d}{x}\right.}=\frac{{x}^{{{n}+{1}}}}{{{n}+{1}}}+{C}{\left({n}\ne-{1}\right)} \displaystyle\int{x}^{{5}}{\left.{d}{x}\right.}=\frac{{x}^{{6}}}{{6}}+{C}

BTW, and assuming \;x>0\;: \lim_{a\to-1}\frac{x^{a+1}-1}{a+1}\stackrel{l'H}=\lim_{a\to-1}x^{a+1}\log x=\log x\neq 0\;\;,\;\;\text{provided}\;\;x\neq 1 so that you in fact get the same in both. ...

Another thought. If 0<a<b, n\ne -1, then \int_a^b x^n \;dx = \frac{b^{n+1}-a^{n+1}}{n+1} and the limit of that as n \to -1 is \ln b - \ln a. Not infinity. So, in a certain sense, your ...