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Question-211979




Question Number 211979 by Spillover last updated on 25/Sep/24
Answered by BHOOPENDRA last updated on 25/Sep/24
∫(dx/((x^2 tan^(−1) x+tan^(−1) x +x^2 π+π)))  ∫(dx/((x^2 +1)(tan^(−1) x+π)))  Let  t=tan^(−1) x+π           dt=(dx/(1+x^2 ))  ∫ (1/t)dt= ln(t)   put t=(tan^(−1) x +π)  ln(tan^(−1) x+π)+c
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} {x}+\mathrm{tan}^{−\mathrm{1}} {x}\:+{x}^{\mathrm{2}} \pi+\pi\right)} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{tan}^{−\mathrm{1}} {x}+\pi\right)} \\ $$$${Let}\:\:{t}=\mathrm{tan}^{−\mathrm{1}} {x}+\pi \\ $$$$\:\:\:\:\:\:\:\:\:{dt}=\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\int\:\frac{\mathrm{1}}{{t}}{dt}=\:{ln}\left({t}\right)\: \\ $$$${put}\:{t}=\left(\mathrm{tan}^{−\mathrm{1}} {x}\:+\pi\right) \\ $$$${ln}\left(\mathrm{tan}^{−\mathrm{1}} {x}+\pi\right)+{c} \\ $$
Commented by Spillover last updated on 26/Sep/24
great.thanks
$${great}.{thanks} \\ $$

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