If you want a way to proceed, we can make use of the Taylor expansion of the arctangent function: \arctan(x) = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{2k+1}\ x^{2k+1} Hence you get: \int e^{2\left(\sum_{k = 0}^{+\infty} \frac{(-1)^k}{2k+1}\ x^{2k+1}\right)}\ \text{d}x ...

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How do you differentiate \displaystyle{g{{\left({x}\right)}}}={e}^{{{1}-{x}}}{\tan{{x}}} using the product rule?
\displaystyle{g}'{\left({x}\right)}={e}^{{{1}-{x}}}{{\sec}^{{2}}{x}}-{e}^{{{1}-{x}}}{\tan{{x}}} Explanation:let\displaystyle{f}={e}^{{{1}-{x}}}and\displaystyle{g}={\tan{{x}}}\displaystyle{f}'={e}^{{{1}-{x}}}\cdot-{1},{g}'={{\sec}^{{2}}{x}} ...
\displaystyle\frac{{e}^{{x}}}{{{1}+{e}^{{{2}{x}}}}} Explanation:The derivative of\displaystyle{\arctan{{\left({x}\right)}}}is: \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{\left({\arctan{{\left({x}\right)}}}\right)}=\frac{{1}}{{{1}+{x}^{{2}}}} ...
\displaystyle\frac{{1}}{{{1}+{x}^{{2}}}}+{4}{e}^{{x}} Explanation: \displaystyle•{\left({x}\right)}\frac{{d}}{{\left.{d}{x}\right.}}{\left({\arctan{{x}}}\right)}=\frac{{1}}{{{1}+{x}^{{2}}}}\ \text{ and }\ \frac{{d}}{{\left.{d}{x}\right.}}{\left({e}^{{x}}\right)}={e}^{{x}} ...

Xem thêm: Nguyễn Cao Sơn Thạch Cao Sơn Nguyễn, Tiểu Sử Của Ca Sĩ S


Integration of this in basic trigonometric functions is not possible, here is the solution in terms hyper-geometric function \int e^{ax} tan(bx) =\\ -\frac{1}{a(a+2ib)} i \left( a e^{(a+2ib)} {}_{2}F_1 \left(1,1-\frac{ia}{2b};2-\frac{ia}{2b};-e^{2ibx}\right) - \\(a+2ib) e^{ax} {}_{2}F_1\left( 1,-\frac{ia}{2b};1-\frac{ia}{2b};-e^{2ibx} \right) \right) + C
How do you differentiate \displaystyle{f{{\left({x}\right)}}}={e}^{{\tan{{\left({x}\right)}}}} using the chain rule?
Multiply the derivative of\displaystyle{e}^{{\tan{{x}}}}by the derivative of\displaystyle{\tan{{x}}}to get\displaystyle{f}'{\left({x}\right)}={e}^{{{\tan{{x}}}}}{{\sec}^{{2}}{x}} ...
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