Could someone please explain how we calculated the first term of the second identity?
Thanks a ton.
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Which term? The $\displaystyle \int \dfrac 1 {(x^2 + 1)^2} , \mathrm dx$ term? – GFauxPas Jul 13 '16 at 11:18
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The term right after it. Sorry for not being clear enough. – Jul 13 '16 at 11:51
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See also https://math.stackexchange.com/questions/35924/integral-of-frac11x22. – Hans Lundmark Mar 17 '21 at 18:26
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Using integration by part with $f'(x)=1$ and $g(x)=\frac{1}{x^2+1}$ yields $$ \int\frac{1}{x^2+1}\,dx =x\cdot\frac{1}{x^2+1}-\int x\cdot\frac{-2x}{(x^2+1)^2}\,dx\\ =\frac{x}{x^2+1}+2\int\frac{x^2}{(x^2+1)^2}\,dx\\ =\frac{x}{x^2+1}+2\int\frac{(x^2+1)-1}{(x^2+1)^2}\,dx\\ =\frac{x}{x^2+1}+2\int\left(\frac{1}{x^2+1}-\frac{1}{(x^2+1)^2}\right)\,dx\\ =\frac{x}{x^2+1}+2\int\frac{1}{x^2+1}\,dx-2\int\frac{1}{(x^2+1)^2}\,dx $$
For the second equality set $$ I=\int\frac{1}{(x^2+1)^2}\,dx\qquad J=\int\frac{1}{x^2+1}\,dx $$ In part 1 we proved that $$ J=\frac{x}{x^2+1}+2J-2I $$ Hence $$ 2I=\frac{x}{x^2+1}+J $$ that is $$ I=\frac{x}{2(x^2+1)}+\frac{J}{2} $$ as requared.
boaz
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Thank you very much. Could you please add how we calculated the second equality? Thanks again! – Jul 13 '16 at 11:50
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