Question Number 55998 by maxmathsup by imad last updated on 07/Mar/19
$${find}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({t}^{\mathrm{2}} +{xt}\:+\mathrm{1}\right){dt}\:\:. \\ $$
Commented by MJS last updated on 08/Mar/19
$$\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{55994}\:\mathrm{but}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{complicated}\:\mathrm{to}\:\mathrm{expand} \\ $$
Answered by MJS last updated on 08/Mar/19
![=∫arctan (t^2 +xt+1) dt= =(1/2)(2t+x)arctan (t^2 +xt+1)−(1/2)∫((√(4t+x^2 −4))/(t^2 +1))dt −(1/2)∫((√(4t+x^2 −4))/(t^2 +1))dt= [u=(√(4t+x^2 −4)) → dt=((√(4t+x^2 −4))/2)du] =−4∫(u^2 /((u−au+b)(u+au+b)))du with a=(√(2x^2 −8+2(√(x^4 −8x^2 +32)))); b=(√(x^4 −8x^2 +32)) now proceed as in 55994](https://www.tinkutara.com/question/Q56013.png)
$$=\int\mathrm{arctan}\:\left({t}^{\mathrm{2}} +{xt}+\mathrm{1}\right)\:{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{t}+{x}\right)\mathrm{arctan}\:\left({t}^{\mathrm{2}} +{xt}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{\mathrm{4}{t}+{x}^{\mathrm{2}} −\mathrm{4}}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{\mathrm{4}{t}+{x}^{\mathrm{2}} −\mathrm{4}}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\sqrt{\mathrm{4}{t}+{x}^{\mathrm{2}} −\mathrm{4}}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{4}{t}+{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}{du}\right] \\ $$$$=−\mathrm{4}\int\frac{{u}^{\mathrm{2}} }{\left({u}−{au}+{b}\right)\left({u}+{au}+{b}\right)}{du} \\ $$$$\mathrm{with}\:{a}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}+\mathrm{2}\sqrt{{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{32}}};\:{b}=\sqrt{{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{32}} \\ $$$$\mathrm{now}\:\mathrm{proceed}\:\mathrm{as}\:\mathrm{in}\:\mathrm{55994} \\ $$
Commented by MJS last updated on 08/Mar/19
$$\mathrm{you}'\mathrm{re}\:\mathrm{always}\:\mathrm{welcome} \\ $$
Commented by turbo msup by abdo last updated on 08/Mar/19
$${thanks}\:{sir}\:{MJS}. \\ $$