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

Question

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^{74 x} \arctan (e^{37 x}) d x

Inverse Trigonometric Forms (92):

\int u \tan^{- 1} u d u = \frac{u^{2} + 1}{2} \tan^{- 1} u - \frac{u}{2} + C

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curwyrm

Answered 2021-05-15 Author has 87 answers

Step 1
$\int (37) \cdot e^{74 x} \cdot \tan^{- 1} (e^{37 x}) d x$
$\Rightarrow e^{37 x} = t \Rightarrow (37) e^{37 x} \cdot d x = d t$ $= \int t \cdot \tan^{- 1} (t) d t$
From table given formula number 92:
$\int u \cdot \tan^{- 1} u \cdot d u = \frac{u^{2} + 1}{2} \tan^{- 1} u - \frac{u}{2} + C$
$\Rightarrow \frac{t^{2} + 1}{2} \tan^{- 1} t - \frac{t}{2} + C$
$\Rightarrow \frac{e^{74 x} + 1}{2} \tan^{- 1} (e^{37 x}) - \frac{e^{37 x}}{2} + C$