7.1 Integration By Parts/11: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int \text {arctan(4t)}dt = \text { | \int \text {arctan(4t)}dt = \text {t\cdot arctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u} \\[1ex] | ||
&= \text {tarctan(4t)}-\frac{1}{8}in(u)= \text {tarctan(4t)}-\frac{1}{8}in(1+16t^{2})+C \\[1ex] | &= \text {tarctan(4t)}-\frac{1}{8}in(u)= \text {tarctan(4t)}-\frac{1}{8}in(1+16t^{2})+C \\[1ex] |
Revision as of 05:18, 29 November 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int \text {arctan(4t)}dt = \text {t\cdot arctan(4t)}-4\int \frac{t}{1+16t^{2}} dt = \text {tarctan(4t)}-\frac{4}{32}\int\frac{1}{u} \\[1ex] &= \text {tarctan(4t)}-\frac{1}{8}in(u)= \text {tarctan(4t)}-\frac{1}{8}in(1+16t^{2})+C \\[1ex] & \qquad u=1+16t^{2} \\[1ex] & \qquad du=32t dt \\[1ex] & \qquad \frac{1}{32}du=t dt \end{align} }