From Mr. V Wiki Math
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| \begin{align} | | \begin{align} |
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| \int_{}^{} \left(\ln(x)^{n}\right)dx &= x \ln(x)^{n} - \int_{}^{} \left((x \frac{n \ln(x)^{n-1}}{x}) \right)dx \\[2ex] | | &= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\int_{0}^{1} (-e^{x})(x)dx \\[2ex] |
| &= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex] | | |
| &= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex] | | \end{align} |
| | </math> |
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| | <math> |
| | \begin{align} |
| | u &= x \quad dv= -e^{-x}dx \\[2ex] |
| | du &= dx \qquad v= e^{-x} \\[2ex] |
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| | \end{align} |
| | </math> |
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| | <math> |
| | \begin{align} |
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| | &= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\left[(x)(e^{-x})\right]\bigg|_{0}^{1} - \int_{0}^{1} (e^{-x})dx \\[2ex] |
| | &= -2e^{-1}+1 - \frac{2}{e} + 2(-e^{-1}+e^{0}) \\[2ex] |
| | &= -\frac{2}{e} + 1 - \frac{2}{e} - \frac{2}{e} + 2 \\[2ex] |
| | &= -\frac{6}{e} + 3 |
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| \end{align} | | \end{align} |
| </math> | | </math> |
Latest revision as of 04:02, 29 November 2022
![{\displaystyle {\begin{aligned}u&=x^{2}+1\quad dv=e^{-x}dx\\[2ex]du&=2xdx\qquad v=-e^{-x}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dca2a9cf6138c41cb61d56c1ee91c9c4fea73909)
![{\displaystyle {\begin{aligned}&=(x^{2}+1)(-e^{-x}){\bigg |}_{0}^{1}-2\int _{0}^{1}(-e^{x})(x)dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/06474884d0ba6d6fc8fae829d72953453d786c3a)
![{\displaystyle {\begin{aligned}u&=x\quad dv=-e^{-x}dx\\[2ex]du&=dx\qquad v=e^{-x}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cfe5f1427cf8d21f06f446c8bf4b0a6dc88f5acf)
![{\displaystyle {\begin{aligned}&=(x^{2}+1)(-e^{-x}){\bigg |}_{0}^{1}-2\left[(x)(e^{-x})\right]{\bigg |}_{0}^{1}-\int _{0}^{1}(e^{-x})dx\\[2ex]&=-2e^{-1}+1-{\frac {2}{e}}+2(-e^{-1}+e^{0})\\[2ex]&=-{\frac {2}{e}}+1-{\frac {2}{e}}-{\frac {2}{e}}+2\\[2ex]&=-{\frac {6}{e}}+3\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/815ca1b063dc1163ade31dbb4c2353e260d3c54e)