From Mr. V Wiki Math
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| <math> | | <math> |
| du= \frac{4}{1+(4t)^{2}} dt \qquad v=t | | du= \frac{4}{1+(4t)^{2}} dt \qquad v=t |
| | </math> |
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| | <math> |
| | \int \text {arctan(4t)}dt = t \cdot \text {arctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = t \cdot \text {arctan(4t)}-\frac{4}{32}\int\frac{1}{u} = t \cdot \text {arctan(4t)}-\frac{1}{8}\ln(u)= t \cdot \text {arctan(4t)}-\frac{1}{8}\ln(1+16t^{2})+C |
| </math> | | </math> |
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| <math> | | <math> |
| \begin{align} | | \begin{align} |
| \int \text {arctan(4t)}dt = \text {tarctan(4t)}-4\int \frac{t}{1+16t^{2}} dt\\[2ex]
| | & u=1+16t^{2} \\ [1ex] |
| | | & du=32t dt \\ [1ex] |
| &= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex] | | & \frac{1}{32}du=t dt |
| &= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex] | |
| &= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex]
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| &= \pi\left[4+\frac{8}{12}-3-\frac{1}{12}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex]
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| &= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex]
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| &= \frac{19\pi}{12}
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| \end{align} | | \end{align} |
| </math>
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| <math>\int (xe^{-x}) dx = x*(-e^{-x}) - \int (-e^{-x})
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| </math> | | </math> |
Latest revision as of 05:55, 29 November 2022
![{\displaystyle {\begin{aligned}&u=1+16t^{2}\\[1ex]&du=32tdt\\[1ex]&{\frac {1}{32}}du=tdt\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ce51b65e9c6d3b8c392f670280a4487d8c5469c7)