Use the following integration formula for inverse trigonometric functions. du 1 arctan ) + C a² + u? a a Therefore, 8 1 -arctan + C I (1)2 + (8x)² %3D E = xp 1 x) + c = 3 arctan
Use the following integration formula for inverse trigonometric functions. du 1 arctan ) + C a² + u? a a Therefore, 8 1 -arctan + C I (1)2 + (8x)² %3D E = xp 1 x) + c = 3 arctan
Calculus: Early Transcendentals
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Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Use the following integration formula for inverse trigonometric functions.
\[
\int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan \left( \frac{u}{a} \right) + C
\]
Therefore,
\[
3 \int \frac{8}{(1)^2 + (8x)^2} \, dx = 3 \left( \frac{1}{8} \arctan \left( \frac{\boxed{8}}{\boxed{1}} x \right) \right) + C
\]
\[
= 3 \arctan \left( \boxed{8} x \right) + C.
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25721e20-facb-4d0d-8f2d-fb325bdcbfad%2Febde67e9-fdde-4c10-901e-87948f5612ad%2Fkkrjykr_processed.png&w=3840&q=75)
Transcribed Image Text:Use the following integration formula for inverse trigonometric functions.
\[
\int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan \left( \frac{u}{a} \right) + C
\]
Therefore,
\[
3 \int \frac{8}{(1)^2 + (8x)^2} \, dx = 3 \left( \frac{1}{8} \arctan \left( \frac{\boxed{8}}{\boxed{1}} x \right) \right) + C
\]
\[
= 3 \arctan \left( \boxed{8} x \right) + C.
\]
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