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2022-01-25Added 13 answers
I=∫0∞xsinx(1+x2)3dx=−14sinx(1+x2)2∣0∞⏟=0+14∫0∞cosx(1+x2)2dxNow we have that I=−14J′(1) because we can take:J(a)=∫0∞cosxa+x2dx⇒J′(a)=−∫0∞cosx(a+x2)2dxThus all we need to do is to find J(a) then take a derivate and set a=1.J(a)=π2ae−a⇒J′(a)=π2(−12a−32e−a−1ae−a12a)⇒I=−14J′(1)=π8(12e−1+12e−1)=π8e
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