Question Number 84708 by Power last updated on 15/Mar/20
Answered by TANMAY PANACEA last updated on 16/Mar/20
$$\int\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{5}}} \\ $$$$\int\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}}} \\ $$$${x}+\mathrm{2}=\frac{\mathrm{1}}{{t}}\rightarrow{dx}=\frac{−{dt}}{{t}^{\mathrm{2}} } \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\sqrt{\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}}−\mathrm{5}}} \\ $$$$\int\frac{−{t}\:{dt}}{\:\sqrt{\mathrm{1}−\mathrm{2}{t}−\mathrm{5}{t}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}\int\frac{−\mathrm{10}{t}−\mathrm{2}+\mathrm{2}}{\:\sqrt{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}}{dt} \\ $$$${now}… \\ $$$$−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1} \\ $$$$\mathrm{5}\left\{\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{5}}\right)^{\mathrm{2}} −\left({t}+\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} \right\} \\ $$$$\boldsymbol{{so}} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}\int\frac{{d}\left(−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)}{\:\sqrt{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dt}}{\:\sqrt{\mathrm{5}}\:\sqrt{\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{5}}\right)^{\mathrm{2}} −\left({t}+\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} }\:} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}×\frac{\left(−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{5}}}{sin}^{−\mathrm{1}} \left(\frac{{t}+\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\sqrt{\mathrm{6}}}{\mathrm{5}}}\right)+\boldsymbol{{C}} \\ $$$${pls}\:{replace}\:{t}\:{by}\:\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$ \\ $$
Commented by Power last updated on 15/Mar/20
$$\mathrm{thanks} \\ $$
Commented by TANMAY PANACEA last updated on 15/Mar/20
$${most}\:{welcome} \\ $$