Question Number 49367 by behi83417@gmail.com last updated on 06/Dec/18
$$\left.\:\:\:\:{a}\right)\:\:\int\:\:\frac{\boldsymbol{\mathrm{dx}}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{tgx}}}} \\ $$$$\left.\:\:\:\:{b}\right)\int\:\:\frac{\boldsymbol{\mathrm{dx}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\boldsymbol{\mathrm{tgx}}}} \\ $$$$\left.\:\:\:\:{c}\right)\int\:\:\frac{\boldsymbol{\mathrm{dx}}}{\:\sqrt{\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}}}} \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${thank}\:{you}\:{very}\:{much}\:{pro}.\:{Abdo}. \\ $$
Commented by maxmathsup by imad last updated on 06/Dec/18
$${you}\:{are}\:{welcome}. \\ $$
Commented by maxmathsup by imad last updated on 06/Dec/18
$$\left.{c}\right){trigonometric}\:{method}\:\:{changement}\:{x}={sin}^{\mathrm{2}} {t}\:{give} \\ $$$$\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}}\:=\:\int\:\:\frac{\mathrm{2}{sint}\:{cost}}{\:\sqrt{\mathrm{1}−{cost}}}\:{dt}\:=\mathrm{2}\:\int\:\:\frac{{sint}\:{cost}}{\:\sqrt{\mathrm{2}}{sin}\left(\frac{{t}}{\mathrm{2}}\right)}{dt} \\ $$$$=\mathrm{2}\:\int\:\:\frac{\mathrm{2}{sin}\left(\frac{{t}}{\mathrm{2}}\right){cos}\left(\frac{{t}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}{sin}\left(\frac{{t}}{\mathrm{2}}\right)}\:\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)\right){dt} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}\:\int\:\left({cos}\left(\frac{{t}}{\mathrm{2}}\right)−\mathrm{2}\:{cos}\left(\frac{{t}}{\mathrm{2}}\right){sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)\right){dt} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}\:\int\:{cos}\left(\frac{{t}}{\mathrm{2}}\right){dt}\:−\frac{\mathrm{8}}{\:\sqrt{\mathrm{2}}}\:\int\:{cos}\left(\frac{{t}}{\mathrm{2}}\right){sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right){dt} \\ $$$$=\frac{\mathrm{8}}{\:\sqrt{\mathrm{2}}}{sin}\left(\frac{{t}}{\mathrm{2}}\right)\:−\frac{\mathrm{16}}{\mathrm{3}\sqrt{\mathrm{2}}}\:{sin}^{\mathrm{3}} \left(\frac{{t}}{\mathrm{2}}\right)+{c} \\ $$$$=\frac{\mathrm{8}}{\:\sqrt{\mathrm{2}}}{sin}\left(\frac{{arcsin}\left(\sqrt{{x}}\right)}{\mathrm{2}}\right)−\frac{\mathrm{16}}{\mathrm{3}\sqrt{\mathrm{2}}}\:{sin}^{\mathrm{3}} \left(\frac{{arcsin}\left(\sqrt{{x}}\right)}{\mathrm{2}}\right)+{c}\:. \\ $$
Commented by maxmathsup by imad last updated on 06/Dec/18
$$\left.{c}\right)\:{let}\:{I}\:=\int\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}}\:\:{changement}\:\sqrt{\mathrm{1}−{x}}={t}\:{give}\:\mathrm{1}−{x}={t}^{\mathrm{2}} \:{and} \\ $$$${I}\:=\:\int\:\:\:\frac{−\mathrm{2}{tdt}}{\:\sqrt{\mathrm{1}−{t}}}\:=\mathrm{2}\:\int\:\:\frac{\mathrm{1}−{t}\:−\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}}}{dt}\:=\mathrm{2}\:\int\sqrt{\mathrm{1}−{t}}{dt}\:−\mathrm{2}\:\int\:\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}}}\:{but} \\ $$$$\int\:\:\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}}}\:=−\mathrm{2}\sqrt{\mathrm{1}−{t}}\:\:{and}\:\int\:\sqrt{\mathrm{1}−{t}}{dt}\:=\int\:\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:=−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\mathrm{4}\sqrt{\mathrm{1}−{t}}+{c}\:=−\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{4}\sqrt{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}+{c}\:. \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${nice}\:{method}.{i}\:{love}\:{this}.{thanks}\:{a}\:{lot}. \\ $$
Answered by MJS last updated on 06/Dec/18
$$\left({a}\right) \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}\right] \\ $$$$=−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}=−\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{t}+\sqrt{\mathrm{2}}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{t}+\sqrt{\mathrm{2}}\right)}= \\ $$$$=−\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −{pt}+{q}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right)}= \\ $$$$=\frac{\mathrm{1}}{{pq}}\int\frac{{t}−{p}}{{t}^{\mathrm{2}} −{pt}+{q}}{dt}−\frac{\mathrm{1}}{{pq}}\int\frac{{t}+{p}}{{t}^{\mathrm{2}} +{pt}+{q}}{dt} \\ $$$$\mathrm{and}\:\mathrm{these}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
$$\left.{a}\right)\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{tanx}}}\:\:\: \\ $$$${t}^{\mathrm{2}} =\mathrm{1}−{tanx}\:\:\mathrm{2}{tdt}=−{sec}^{\mathrm{2}} {xdx} \\ $$$${dx}=\frac{−\mathrm{2}{tdt}}{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\int\frac{−\mathrm{2}{tdt}}{{t}×\left\{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \right\}} \\ $$$$=−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}} \\ $$$$=−\mathrm{2}\int\frac{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{dt} \\ $$$${now}\:{t}^{\mathrm{2}} +\frac{\mathrm{2}}{{t}^{\mathrm{2}} }=\left({t}+\frac{\sqrt{\mathrm{2}}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({t}−\frac{\sqrt{\mathrm{2}}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$${d}\left({t}+\frac{\sqrt{\mathrm{2}}}{{t}}\right)=\left(\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }\right){dt} \\ $$$${d}\left({t}−\frac{\sqrt{\mathrm{2}}}{{t}}\right)=\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }\right)}{{t}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\int\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\sqrt{\mathrm{2}}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}}{dt}−\int\frac{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\sqrt{\mathrm{2}}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}}{dt}\right] \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\sqrt{\mathrm{2}}}{{t}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}{ln}\left\{\frac{\left({t}+\frac{\sqrt{\mathrm{2}}}{{t}}\right)−\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}{\left({t}+\frac{\sqrt{\mathrm{2}}}{{t}}\right)+\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}\right\}+{c}\right. \\ $$$${now}\:{pls}\:{put}\:{t}=\sqrt{\mathrm{1}−{tanx}}\: \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${thanks}\:{in}\:{advance}\:{sir}\:{MJS}\:{and}\:{sir}\:{tanmay}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
$$\left.{c}\right){t}^{\mathrm{2}} =\mathrm{1}−\sqrt{\mathrm{1}−{x}}\: \\ $$$$\left(\mathrm{1}−{x}\right)=\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{1}−{x}=\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \\ $$$${x}={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:\: \\ $$$${dx}=\left(\mathrm{4}{t}^{\mathrm{3}} −\mathrm{4}{t}\right){dt} \\ $$$$\int\frac{\mathrm{4}{t}^{\mathrm{3}} −\mathrm{4}{t}}{{t}}{dt} \\ $$$$\mathrm{4}\int{t}^{\mathrm{2}} {dt}−\mathrm{4}\int{dt} \\ $$$$=\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{4}{t}+{c} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left\{\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\:\right)\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{c} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
$${trying}\:{to}\:{solve}\:{q}.{no}\:\left({b}\right)…{excellent}\:{posts}\:{from}\:{your}\:{side}\:{sir}.. \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${you}\:{are}\:{wellcome}\:{sir}.{one}\:{of}\:\:{the}\:{best}\: \\ $$$${question}\:{solvers}\:{with}\:{great}\:{works} \\ $$$${in}\:{this}\:{lovely}\:{forum}\:{is}:{sir}\:{tanmay}. \\ $$$${godlock}\:{dear}. \\ $$
Answered by MJS last updated on 06/Dec/18
$$\left({b}\right) \\ $$$$\int\frac{{dx}}{\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }= \\ $$$$\:\:\:\:\:\left[{t}=\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\rightarrow\:{dx}=−\frac{\mathrm{3}{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}}\right] \\ $$$$=−\mathrm{3}\int\frac{{t}}{{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}}{dt} \\ $$$${t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}=\left({t}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{\mathrm{4}}{t}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)\left({t}^{\mathrm{2}} −\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}{t}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)\left({t}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}{t}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)= \\ $$$$=\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} +{rt}+{q}\right)\left({t}^{\mathrm{2}} +{st}+{q}\right)\:\Rightarrow \\ $$$$\Rightarrow\:\frac{\mathrm{3}}{{q}\left({p}−{r}\right)\left({p}−{s}\right)}\int\frac{{t}+{p}}{{t}^{\mathrm{2}} +{pt}+{q}}{dt}+\frac{\mathrm{3}}{{q}\left({r}−{p}\right)\left({r}−{s}\right)}\int\frac{{t}+{r}}{{t}^{\mathrm{2}} +{rt}+{q}}{dt}+\frac{\mathrm{3}}{{q}\left({s}−{p}\right)\left({s}−{r}\right)}\int\frac{{t}+{s}}{{t}^{\mathrm{2}} +{st}+{q}}{dt} \\ $$$$…\mathrm{and}\:\mathrm{again}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{these} \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${ok}!\:{thanks}.\:{waiting}….. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$$\left.{b}\right){t}^{\mathrm{3}} =\mathrm{1}−{tanx}\: \\ $$$$\mathrm{3}{t}^{\mathrm{2}} {dt}=−{sec}^{\mathrm{2}} {xdx} \\ $$$${dx}=\frac{−\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{3}} \right)^{\mathrm{2}} }{dt}=\frac{−\mathrm{3}{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}}{dt} \\ $$$$\int\frac{−\mathrm{3}{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}}×\frac{\mathrm{1}}{{t}}{dt} \\ $$$$=−\mathrm{3}\int\frac{{tdt}}{{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}} \\ $$$${t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2} \\ $$$$=\left({t}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}×{t}^{\mathrm{3}} ×\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$=−\mathrm{3}\int\frac{{tdt}}{\left({t}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} −{i}^{\mathrm{2}} } \\ $$$$=−\mathrm{3}\int\frac{{tdt}}{\left({t}^{\mathrm{3}} −\mathrm{1}−{i}\right)\left({t}^{\mathrm{3}} −\mathrm{1}+{i}\right)} \\ $$$$=−\mathrm{3}\int\frac{{tdt}}{\left({t}^{\mathrm{3}} +{a}\right)\left({t}^{\mathrm{3}} +{b}\right)}\:\left[{a}=−\mathrm{1}−{i}\:\:\:\:{b}=−\mathrm{1}+{i}\right] \\ $$$$=\frac{−\mathrm{3}}{{a}−{b}}\int\left[\frac{{t}\left({t}^{\mathrm{3}} +{a}\right)−{t}\left({t}^{\mathrm{3}} +{b}\right)}{\left({t}^{\mathrm{3}} +{a}\right)\left({t}^{\mathrm{3}} +{b}\right)}\right]{dt} \\ $$$$=\frac{\mathrm{3}}{{b}−{a}}\left[\int\frac{{tdt}}{{t}^{\mathrm{3}} +{b}}−\int\frac{{tdt}}{{t}^{\mathrm{3}} +{a}}\right] \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{solving}}\:\int\frac{{tdt}}{{t}^{\mathrm{3}} +{a}} \\ $$$$\int\frac{{tdt}}{\left({t}+^{\mathrm{3}} \sqrt{{a}}\:\right)\left({t}^{\mathrm{2}} −{t}×^{\mathrm{3}} \sqrt{{a}}\:+^{\frac{\mathrm{2}}{\mathrm{3}}} \sqrt{{a}}\:\right)} \\ $$$$\int\frac{{tdt}}{\left({t}+{p}\right)\left({t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} \right)} \\ $$$$\frac{{t}}{\left({t}+{p}\right)\left({t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} \right)}=\frac{{A}}{{t}+{p}}+\frac{{Bt}+{C}}{{t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} } \\ $$$${t}={A}\left({t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} \right)+\left({Bt}+{C}\right)\left({t}+{p}\right) \\ $$$${t}={t}^{\mathrm{2}} \left({A}+{B}\right)+{t}\left(−{Ap}+{Bp}+{C}\right)+{Ap}^{\mathrm{2}} +{Cp} \\ $$$${A}+{B}=\mathrm{0} \\ $$$${Ap}^{\mathrm{2}} +{Cp}=\mathrm{0} \\ $$$$−{Ap}+{Bp}+{C}=\mathrm{1} \\ $$$${C}=−{Ap} \\ $$$$−{Ap}−{Ap}−{Ap}=\mathrm{1}\:\:\:\:{A}=\frac{−\mathrm{1}}{\mathrm{3}{p}}\:\:\:{B}=\frac{\mathrm{1}}{\mathrm{3}{p}}\:\:\:\:{C}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\int\frac{{A}}{{t}+{p}}{dt}+\int\frac{{Bt}+{C}}{{t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} }{dt} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}{p}}\int\frac{{dt}}{{t}+{p}}+\int\frac{\frac{\mathrm{1}}{\mathrm{3}{p}}{t}+\frac{−\mathrm{1}}{\mathrm{3}}}{{t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} }{dt} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}{p}}\int\frac{{dt}}{{t}+{p}}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} }{dt} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}{p}}\int\frac{{dt}}{{t}+{p}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}−{p}+{p}−\mathrm{2}}{{t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} } \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}{p}}\int\frac{{dt}}{{t}+{p}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{d}\left({t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} }+\frac{{p}−\mathrm{2}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}.{t}.\frac{{p}}{\mathrm{2}}+\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{p}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}{p}}{ln}\left({t}+{p}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({t}^{\mathrm{2}} −{tp}+{p}^{\mathrm{2}} \right)+\frac{{p}−\mathrm{2}}{\mathrm{6}}×\frac{\mathrm{1}}{\left(\frac{\sqrt{\mathrm{3}}\:{p}}{\mathrm{2}}\right)^{} }×{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{{p}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}\:{p}}{\mathrm{2}}}\right)+{c}_{\mathrm{1}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{put}}\:\boldsymbol{{p}}=\left(\boldsymbol{{a}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \: \\ $$$$\boldsymbol{{next}}\:\boldsymbol{{put}}\:\boldsymbol{{a}}=\left(−\mathrm{1}−\boldsymbol{{i}}\right) \\ $$$$\boldsymbol{{similarly}}\:\boldsymbol{{we}}\:\boldsymbol{{can}}\:\boldsymbol{{find}}\:\int\frac{{tdt}}{{t}^{\mathrm{3}} +{b}} \\ $$$${in}\:{place}\:{of}\:{p}\:{put}\:\left({b}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${next}\:{put}\:{b}=\left(−\mathrm{1}+{i}\right) \\ $$$${finaly}\:{algebric}\:{addition}… \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 07/Dec/18
$${great}\:{work}\:{done}\:{by}\:{you}.{thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${thank}\:{you}\:{sir}… \\ $$