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Question Number 166141 by Lambert last updated on 13/Feb/22

$$\int_{\mathrm{0}} ^{\boldsymbol{\mathrm{x}}} \frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}\boldsymbol{\mathrm{dt}}\: \\$$

Answered by MJS_new last updated on 14/Feb/22

$$\int\frac{{t}^{\mathrm{2}} }{\:\sqrt{{a}+\mathrm{2}{t}^{\mathrm{2}} }}{dt}= \\$$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{{a}}}\left({t}+\sqrt{{a}+\mathrm{2}{t}^{\mathrm{2}} }\right)\:\rightarrow\:{dt}=\frac{\sqrt{{a}+\mathrm{2}{t}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}}{u}}{du}\right] \\$$$$=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{16}}\int\frac{\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{3}} }{du}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{16}}\int\left({u}−\frac{\mathrm{2}}{{u}}+\frac{\mathrm{1}}{{u}^{\mathrm{3}} }\right){du}= \\$$$$=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{16}}\left(\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2ln}\:{u}\:−\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }\right)=... \\$$$$=\frac{{t}\sqrt{{a}+\mathrm{2}{t}^{\mathrm{2}} }}{\mathrm{4}}−\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}{t}+\sqrt{{a}+\mathrm{2}{t}^{\mathrm{2}} }\right)\:+{C} \\$$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\$$

Commented by cortano1 last updated on 14/Feb/22

$$\:\mathrm{u}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{a}}}\:\mathrm{t}?\:\mathrm{sir} \\$$

Commented by MJS_new last updated on 14/Feb/22

$$\mathrm{typo}\:=\:\mathrm{instead}\:\mathrm{of}\:+;\:\mathrm{it}'\mathrm{s}\:\mathrm{now}\:\mathrm{corrected} \\$$

Answered by Mathspace last updated on 14/Feb/22

$${f}\left({a}\right)=\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{t}^{\mathrm{2}} }{{a}}}}{dt} \\$$$$=\frac{{a}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{{x}} \:\frac{\frac{\mathrm{2}}{{a}}{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{t}^{\mathrm{2}} }{{a}}}}{dt} \\$$$$=\frac{\sqrt{{a}}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \sqrt{\mathrm{1}+\frac{\mathrm{2}{t}^{\mathrm{2}} }{{a}}}{dt}−\frac{\sqrt{{a}}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{t}^{\mathrm{2}} }{{a}}}} \\$$$$\frac{\sqrt{\mathrm{2}}{t}}{\:\sqrt{{a}}}={shu}\:\Rightarrow{u}={argsh}\left(\frac{\sqrt{\mathrm{2}}{t}}{\:\sqrt{{a}}}\right) \\$$$$\int_{\mathrm{0}} ^{{x}} \sqrt{\mathrm{1}+\frac{\mathrm{2}{t}^{\mathrm{2}} }{{a}}}{dt}=\int_{\mathrm{0}} ^{{argsh}\left(\frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{{a}}}\right)} {chu}.\frac{\sqrt{{a}}}{\:\sqrt{\mathrm{2}}}{chudu} \\$$$$=\frac{\sqrt{{a}}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{{ln}\left(\frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{{a}}}+\sqrt{\mathrm{1}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{{a}}}\right)} \:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{\mathrm{2}}{du} \\$$$$=\frac{\sqrt{{a}}}{\mathrm{2}}{ln}\left(....\right)+\frac{\sqrt{{a}}}{\mathrm{4}\sqrt{\mathrm{2}}}\left[{sh}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{{ln}\left(...\right)} \\$$$$=\frac{\sqrt{{a}}}{\mathrm{2}}{ln}\left(...\right)+\frac{\sqrt{{a}}}{\mathrm{4}\sqrt{\mathrm{2}}}\left[\frac{{e}^{\mathrm{2}{u}} −{e}^{−\mathrm{2}{u}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(...\right)} \\$$$$=\frac{\sqrt{{a}}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{{a}}}+\sqrt{\mathrm{1}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{{a}}}\right) \\$$$$\frac{\sqrt{{a}}}{\mathrm{8}\sqrt{\mathrm{2}}}\left\{\:\left(\frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{{a}}}+\sqrt{\mathrm{1}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{{a}}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{{a}}}+\sqrt{\mathrm{1}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{{a}}}\right)^{−\mathrm{2}} \right\} \\$$$${same}\:{way}\:{for}\:{other}\:{integral}... \\$$