Answer
$$\frac{\pi}{4}$$
Work Step by Step
Given
$$\int_0^1\frac{dt}{1+t^2} $$
Since
\begin{aligned}
\int_0^1\frac{dt}{1+t^2}&=\tan^{-1}(t)\bigg|_{0}^{1}\\
&= \tan^{-1}(1)-\tan^{-1}(0)\\
&=\frac{\pi}{4}
\end{aligned}
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 330: 82
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