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Question Number 124742 by TANMAY PANACEA last updated on 05/Dec/20

∫((3^t +11)/(6^t +11))dt collected problem

$$\int\frac{\mathrm{3}^{{t}} +\mathrm{11}}{\mathrm{6}^{{t}} +\mathrm{11}}{dt}\:\:\:\boldsymbol{{collected}}\:\boldsymbol{{problem}} \\ $$

Commented by MJS_new last updated on 05/Dec/20

I get to this point: ((3^t +11)/(6^t +11))=(1/2^t )+((11)/(6^t +11))−((11)/(2^t (6^t +11)))= =(1/2^t )+((6^t +11)/(6^t +11))−(6^t /(6^t +11))−((11)/(2^t (6^t +11)))= =(1/2^t )+1−(6^t /(6^t +11))−((11)/(2^t (6^t +11))) ∫(dt/2^t )+∫dt−∫(6^t /(6^t +11))dt=−(1/(2^t ln 2))+t−((ln (6^t +11))/(ln 6)) but no idea for −11∫(dt/(2^t (6^t +11)))

$$\mathrm{I}\:\mathrm{get}\:\mathrm{to}\:\mathrm{this}\:\mathrm{point}: \\ $$$$\frac{\mathrm{3}^{{t}} +\mathrm{11}}{\mathrm{6}^{{t}} +\mathrm{11}}=\frac{\mathrm{1}}{\mathrm{2}^{{t}} }+\frac{\mathrm{11}}{\mathrm{6}^{{t}} +\mathrm{11}}−\frac{\mathrm{11}}{\mathrm{2}^{{t}} \left(\mathrm{6}^{{t}} +\mathrm{11}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{t}} }+\frac{\mathrm{6}^{{t}} +\mathrm{11}}{\mathrm{6}^{{t}} +\mathrm{11}}−\frac{\mathrm{6}^{{t}} }{\mathrm{6}^{{t}} +\mathrm{11}}−\frac{\mathrm{11}}{\mathrm{2}^{{t}} \left(\mathrm{6}^{{t}} +\mathrm{11}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{t}} }+\mathrm{1}−\frac{\mathrm{6}^{{t}} }{\mathrm{6}^{{t}} +\mathrm{11}}−\frac{\mathrm{11}}{\mathrm{2}^{{t}} \left(\mathrm{6}^{{t}} +\mathrm{11}\right)} \\ $$$$\int\frac{{dt}}{\mathrm{2}^{{t}} }+\int{dt}−\int\frac{\mathrm{6}^{{t}} }{\mathrm{6}^{{t}} +\mathrm{11}}{dt}=−\frac{\mathrm{1}}{\mathrm{2}^{{t}} \:\mathrm{ln}\:\mathrm{2}}+{t}−\frac{\mathrm{ln}\:\left(\mathrm{6}^{{t}} +\mathrm{11}\right)}{\mathrm{ln}\:\mathrm{6}} \\ $$$$\mathrm{but}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{for}\:−\mathrm{11}\int\frac{{dt}}{\mathrm{2}^{{t}} \left(\mathrm{6}^{{t}} +\mathrm{11}\right)} \\ $$

Commented by MJS_new last updated on 05/Dec/20

maybe the hypergeometric function is possible but I′m not good at using it...

$$\mathrm{maybe}\:\mathrm{the}\:\mathrm{hypergeometric}\:\mathrm{function}\:\mathrm{is} \\ $$$$\mathrm{possible}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{good}\:\mathrm{at}\:\mathrm{using}\:\mathrm{it}... \\ $$

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