From Mr. V Wiki 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 = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u}\\ [1ex] | | \int \text {arctan(4t)}dt = \text {tarctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u} = \text {tarctan(4t)}-\frac{1}{8}in(u)= \text {tarctan(4t)}-\frac{1}{8}in(1+16t^{2})+C \\ [1ex] |
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| u=1+16t^{2} \\[1ex] | | u=1+16t^{2} \\[1ex] |
| du=32t dt \\[1ex] | | du=32t dt \\[1ex] |
| \frac{1}{32}du=t dt | | \frac{1}{32}du=t dt |
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| \end{align}
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| </math>
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| &= \text {tarctan(4t)} \\[1ex]
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| &= \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]
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| &= \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> | | </math> |
Revision as of 05:07, 29 November 2022
![{\displaystyle {\begin{aligned}\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt={\text{tarctan(4t)}}-{\frac {4}{32}}\int {\frac {1}{u}}={\text{tarctan(4t)}}-{\frac {1}{8}}in(u)={\text{tarctan(4t)}}-{\frac {1}{8}}in(1+16t^{2})+C\\[1ex]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/b6103242ea8924fe4c32867f555f2d13d2166f71)