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\(\int x^{2020}\left(\tan ^{-1} x+\cot ^{-1} x\right) d x=\)
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Verified Answer
The correct answer is:
\(\frac{x^{2021}}{2021}\left(\tan ^{-1} x+\cot ^{-1} x\right)+c\)
\(\begin{aligned}
I & =\int x^{2020}\left(\tan ^{-1} x+\cot ^{-1} x\right) d x \\
& =\int x^{2020}\left(\frac{\pi}{2}\right) d x \quad\left\{\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right\} \\
& =\frac{\pi}{2} \frac{x^{2021}}{2021}+C=\frac{x^{2021}}{2021}\left(\tan ^{-1} x+\cot ^{-1} x\right)+C
\end{aligned}\)
Hence, option (b) is correct.
I & =\int x^{2020}\left(\tan ^{-1} x+\cot ^{-1} x\right) d x \\
& =\int x^{2020}\left(\frac{\pi}{2}\right) d x \quad\left\{\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right\} \\
& =\frac{\pi}{2} \frac{x^{2021}}{2021}+C=\frac{x^{2021}}{2021}\left(\tan ^{-1} x+\cot ^{-1} x\right)+C
\end{aligned}\)
Hence, option (b) is correct.
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