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Question Number 206224 by mathzup last updated on 09/Apr/24
find∫01arctan(x5)dx
Answered by Frix last updated on 10/Apr/24
∫tan−1xndx=bypartsu′=1∧v=tan−1xn=xtan−1xn−n∫xnx2n+1dxWithn=5andx∈[0,1]weget5ln(1+5)2−(1+5)ln22+5−10(5−5)20π≈.151019250
Answered by mathzup last updated on 10/Apr/24
thankyousir
bypartswegetI=[xarctan(x5)]01−∫01x.5x41+x10dx=π4−5∫01x51+x10dxbut∫01x51+x10dx=∫01x5Σ(−1)nx10ndx=∑n=0∞(−1)n∫01x10n+5dx=∑n=0∞(−1)n×110n+6=∑n=o∞110(2n)+6−∑n=0∞110(2n+1)+6=∑n=0∞(120n+6−120n+16)=10∑n=0∞1(20n+6)(20n+16)=10202∑n=0∞1(n+620)(n+1620)=140×Ψ(1620)−Ψ(620)1620−620=120{Ψ(45)−Ψ(310)}
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