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Question Number 60413 by tanmay last updated on 20/May/19

Commented by Meritguide1234 last updated on 21/May/19

$$\mathrm{why}\mathrm{do}\mathrm{you}\mathrm{post}\mathrm{same}\mathrm{question}\mathrm{and}\mathrm{solition}\mathrm{from}\mathrm{goiit}\mathrm{page}\mathrm{by}\mathrm{Sourav}\mathrm{De}$$

Commented by MJS last updated on 21/May/19

$$\mathrm{why}\mathrm{not}?$$

Commented by tanmay last updated on 21/May/19

$$weareallstudents...wearenotTomashardy$$$$orRamanujam...welearnfromeachother$$$$whatdo[youwanttosay..ijustsharedthesolution$$$$toothers...iamnotinthesideofothers$$$$whothinkthattheyknoweverything...iamstill$$$$astudent...ihavemilestogo..$$

Commented by maxmathsup by imad last updated on 26/May/19

$$thisplatformisopentoallkindofexerciceswedonotdistinguich$$$$thatandthatthankstosirTinkutara....$$$$$$

Answered by tanmay last updated on 21/May/19

$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$$$I={\int }_0^1\frac{{tan}^{-1}\left(\frac{x}{x+1}\right)}{{tan}^{-1}\left(\frac{1+2x-2x^2}{2}\right)}dx$$$$={\int }_0^1\frac{{tan}^{-1}\left(\frac{1-x}{1-x+1}\right)}{{tan}^{-1}\left(\frac{1+2(1-x)-2{(1-x)}^2}{2}\right)}dx$$$$={\int }_0^1\frac{{tan}^{-1}\left(\frac{1-x}{2-x}\right)}{{tan}^{-1}\left(\frac{1+2-2x-2+4x-2x^2}{2}\right)}dx$$$$={\int }_0^1\frac{{tan}^{-1}\left(\frac{1-x}{1+x}\right)}{{tan}^{-1}\left(\frac{1+2x-2x^2}{2}\right)}dx$$$$2I={\int }_0^1\frac{{tan}^{-1}\left(\frac{x}{x+1}\right)+{tan}^{-1}\left(\frac{1-x}{2-x}\right)}{{tan}^{-1}\left(\frac{1+2x-2x^2}{2}\right)}dx$$$$now{N}_r$$$${tan}^{-1}\left(\frac{x}{x+1}\right)+{tan}^{-1}\left(\frac{1-x}{2-x}\right)$$$$={tan}^{-1}\left(\frac{\frac{x}{x+1}+\frac{1-x}{2-x}}{1-\frac{x(1-x)}{(x+1)(2-x)}}\right)$$$$={tan}^{-1}\left(\frac{2x-x^2+1-x^2}{2x-x^2+2-x-x+x^2}\right)$$$$={tan}^{-1}\left(\frac{1+2x-2x^2}{2}\right)=D_r$$$$so\frac{N_r}{D_r}=1$$$$2I={\int }_0^{1}1\times dx$$$$2I=1$$$$I=\frac{1}{2}$$$$$$$$$$

Commented by Meritguide1234 last updated on 21/May/19

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