Question:
How to Evaluate Integral $$\int\;\frac {x^m\ln(x)}{\cos(x)^2}\mathrm{d}x$$
My suggestion $$F(x)=\int\;\frac {x^m\ln(x)}{\cos(x)^2}\mathrm{d}x$$
integration by part
let $$\mathrm{d}u = \frac{1}{\cos(x)^2} \; \; v = x^m\ln(x)\\ \implies u =\tan(x) \; \; \mathrm{d}v = x^{m-1}(m\ln(x)+1)$$
$$F(x)=x^m\ln(x)\tan(x)-\int x^{m-1}\tan(x)(m\ln(x)+1)\mathrm{d}x$$
Help please
A lot of functions do not have an elementary antiderivative. Here, there is no antiderivative expressible with elementary functions, but we can find a series solution. Taylor series for tangent here $$\tan x=\sum_{k=1}^\infty\frac{(-1)^{k+1}4^k(4^k-1)B_{2k}}{(2k)!}x^{2k-1}$$ $$\sec^2 x=\sum_{k=1}^\infty\frac{(-1)^{k-1}4^k(4^k-1)(2k-1)B_{2k}}{(2k)!}x^{2k-2}$$ $$I=\int x^m\ln(x)\sum_{k=1}^\infty\frac{(-1)^{k-1}4^k(4^k-1)(2k-1)B_{2k}}{(2k)!}x^{2k-2}dx$$$$=\sum_{k=1}^\infty\frac{(-1)^{k-1}4^k(4^k-1)(2k-1)B_{2k}}{(2k)!}\int x^{m+2k-2}\ln (x)dx$$$$=\sum_{k=1}^\infty\frac{(-1)^{k-1}4^k(4^k-1)(2k-1)B_{2k}}{(2k)!}\left(\frac{x^{m+2k-1}\ln x}{m+2k-1}-\frac{x^{m+2k-1}}{(m+2k-1)^2}\right)$$ There is no way to continue forth, and no better answer
