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Elan Coil [88]
9 months ago
5

Question attached as screenshot belowPlease help me with my calculus homework I am paying close attention

Mathematics
1 answer:
Wewaii [24]9 months ago
5 0

Given:

\int ^1_{-1}\frac{1}{x^2+1}dx

To evaluate:

As we know,

\int ^{}_{}\frac{1}{x^2+1^{}}dx=\tan ^{-1}(x)+C

So, we have

\begin{gathered} \int ^1_{-1}\frac{1}{x^2+1}dx=\tan ^{-1}(1)-\tan ^{-1}(-1) \\ =\frac{\pi}{4}-(-\frac{\pi}{4}) \\ =\frac{\pi}{4}+\frac{\pi}{4} \\ =\frac{2\pi}{4} \\ =\frac{\pi}{2} \end{gathered}

Hence, the answer is,

\frac{\pi}{2}

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Answer:

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Step-by-step explanation:

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so

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Group like terms

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