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At a first glance, I notice that 11+x2 is the derivative of arctan(x). Meanwhile, partial fraction will not work in this case (at least from my perspective) because the power of the x in 11+x2020 is too large. Then I tried to approach this problem by integration by part. Then I got the following result:
∫∞01(1+x2)(1+x2020)dx=tan−1(x)1+x2020|∞0−∫∞0−2020tan−1(x)x2019(1+x2020)2dx
Since limx→∞tan−1(x)=1 and limx→∞1+x2020=∞
Then
tan−1(x)1+x2020|∞0=0
However, how should I proceed from
∫∞0−2020tan−1(x)x2019(1+x2020)2dx
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