Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.) ∫37e74xarctan⁡(e37x)dx Inverse Trigonometric Forms (92): ∫utan−1u du=u2+12tan−1u−u2+C

Step 1 ∫(37)⋅e74x⋅tan−1(e37x)dx ⇒e37x=t⇒(37)e37x⋅dx=dt =∫t⋅tan−1(t)dt From table given formula number 92: ∫u⋅tan−1u⋅du=u2+12tan−1u−u2+C ⇒t2+12tan−1t−t2+C ⇒e74x+12tan−1(e37x)−e37x2+C

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Calculus and Analysis
Calculus 1
Integrals

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2021-08-17

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

Answer & Explanation

dessinemoie

Skilled2021-08-18Added 90 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:
$\int u \cdot \tan^{- 1} u \cdot d u = \frac{u^{2} + 1}{2} \tan^{- 1} u - \frac{u}{2} + C$
$\Rightarrow \frac{t^{2} + 1}{2} \tan^{- 1} t - \frac{t}{2} + C$
$\Rightarrow \frac{e^{74 x} + 1}{2} \tan^{- 1} (e^{37 x}) - \frac{e^{37 x}}{2} + C$

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