 # Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.) \int37e^{74x}\arctan(e^{37x})dx Inverse Trigonometric Forms (92): \int u\tan^{-1}u\ du=\frac{u^{2}+1}{2}\tan^{-1}u-\frac{u}{2}+C Nann 2021-05-14 Answered
Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)
$\int 37{e}^{74x}\mathrm{arctan}\left({e}^{37x}\right)dx$
Inverse Trigonometric Forms (92):
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Step 1
$\int \left(37\right)\cdot {e}^{74x}\cdot {\mathrm{tan}}^{-1}\left({e}^{37x}\right)dx$
$⇒{e}^{37x}=t⇒\left(37\right){e}^{37x}\cdot dx=dt$$=\int t\cdot {\mathrm{tan}}^{-1}\left(t\right)dt$
From table given formula number 92:
$\int u\cdot {\mathrm{tan}}^{-1}u\cdot du=\frac{{u}^{2}+1}{2}{\mathrm{tan}}^{-1}u-\frac{u}{2}+C$
$⇒\frac{{t}^{2}+1}{2}{\mathrm{tan}}^{-1}t-\frac{t}{2}+C$
$⇒\frac{{e}^{74x}+1}{2}{\mathrm{tan}}^{-1}\left({e}^{37x}\right)-\frac{{e}^{37x}}{2}+C$