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Question Number 47101 by Meritguide1234 last updated on 04/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

∫(dx/((1+x^2 )^n )) =(1/(2(n−1)))×(x/((x^2 +1)^(n−1) ))+((2n−3)/(2n−2))∫(dx/((1+x^2 )^(n−1) )) I_n =∫_0 ^1 (dx/((1+x^2 )^n )) =(1/(2(n−1)))∣(x/((1+x^2 )^(n−1) ))∣_0 ^1 +((2n−3)/(2n−2))I_(n−1) =(1/(2(n−1)))×(1/2^(n−1) )+((2n−3)/(2n−2))I_(n−1) I_n =(1/((n−1)))×(1/2^n )+((2n−3)/(2n−2))I_(n−1) I_1 =∫_0 ^1 (dx/((1+x^2 )^1 ))=(π/4) I_n =(1/((n−1)))×(1/2^n )+((2n−3)/(2n−2))I_(n−1) I_2 =(1/2^2 )+(1/2)×(π/4)=(1/2^2 )+(π/8) I_3 =(1/2)×(1/2^3 )+(3/4)((1/2^2 )+(π/8)) =(1/2^4 )+(3/2^4 )+((3π)/(32)) contd...

$$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}×\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} } \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\mid\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} }\mid_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{1}} }=\frac{\pi}{\mathrm{4}} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\pi}{\mathrm{8}} \\ $$$${I}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\pi}{\mathrm{8}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}\pi}{\mathrm{32}} \\ $$$${contd}... \\ $$$$ \\ $$

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