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Question Number 131527 by Sudip last updated on 05/Feb/21

Answered by Dwaipayan Shikari last updated on 05/Feb/21

(2/7)∫(t/(t+((22)/7)))dt+(5/7)∫(1/(t+((22)/7)))dt =(2/7)t−((44)/(49))log(t+((22)/7))+(5/7)log(t+((22)/7))

$$\frac{\mathrm{2}}{\mathrm{7}}\int\frac{{t}}{{t}+\frac{\mathrm{22}}{\mathrm{7}}}{dt}+\frac{\mathrm{5}}{\mathrm{7}}\int\frac{\mathrm{1}}{{t}+\frac{\mathrm{22}}{\mathrm{7}}}{dt} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}{t}−\frac{\mathrm{44}}{\mathrm{49}}{log}\left({t}+\frac{\mathrm{22}}{\mathrm{7}}\right)+\frac{\mathrm{5}}{\mathrm{7}}{log}\left({t}+\frac{\mathrm{22}}{\mathrm{7}}\right) \\ $$

Answered by mr W last updated on 05/Feb/21

∫((2t+5)/(7t+22))dt =(1/7)∫((2×7t+2×22−9)/(7t+22))dt =(1/7)∫(2−(9/(7t+22)))dt =(1/7)(2t−∫(9/(7t+22))dt) =(1/7)[2t−(9/7)∫(1/(7t+22))d(7t+22)] =(1/7)[2t−(9/7)ln (7t+22)]+C =(2/7)t−(9/(49))ln (7t+22)+C

$$\int\frac{\mathrm{2}{t}+\mathrm{5}}{\mathrm{7}{t}+\mathrm{22}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\int\frac{\mathrm{2}×\mathrm{7}{t}+\mathrm{2}×\mathrm{22}−\mathrm{9}}{\mathrm{7}{t}+\mathrm{22}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\int\left(\mathrm{2}−\frac{\mathrm{9}}{\mathrm{7}{t}+\mathrm{22}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\left(\mathrm{2}{t}−\int\frac{\mathrm{9}}{\mathrm{7}{t}+\mathrm{22}}{dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\left[\mathrm{2}{t}−\frac{\mathrm{9}}{\mathrm{7}}\int\frac{\mathrm{1}}{\mathrm{7}{t}+\mathrm{22}}{d}\left(\mathrm{7}{t}+\mathrm{22}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\left[\mathrm{2}{t}−\frac{\mathrm{9}}{\mathrm{7}}\mathrm{ln}\:\left(\mathrm{7}{t}+\mathrm{22}\right)\right]+{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}{t}−\frac{\mathrm{9}}{\mathrm{49}}\mathrm{ln}\:\left(\mathrm{7}{t}+\mathrm{22}\right)+{C} \\ $$

Answered by Bird last updated on 06/Feb/21

I=∫ ((2t+5)/(7t+22))dt we do the ch.7t+22=x ⇒ t=((x−22)/7) ⇒I=∫ (1/x)(((2x−44)/7)+5)(dx/7) =(1/(49))∫ (1/x)(2x−44+35)dx =(1/(49))∫ ((2x−9)/x) dx =(2/(49))x−(9/(49))ln∣x∣ +C =(2/(49))(7t+22)−(9/(49))ln∣7t+22∣ +C

$${I}=\int\:\frac{\mathrm{2}{t}+\mathrm{5}}{\mathrm{7}{t}+\mathrm{22}}{dt}\:{we}\:{do}\:{the}\:{ch}.\mathrm{7}{t}+\mathrm{22}={x}\:\Rightarrow \\ $$$${t}=\frac{{x}−\mathrm{22}}{\mathrm{7}}\:\Rightarrow{I}=\int\:\:\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{2}{x}−\mathrm{44}}{\mathrm{7}}+\mathrm{5}\right)\frac{{dx}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{49}}\int\:\:\frac{\mathrm{1}}{{x}}\left(\mathrm{2}{x}−\mathrm{44}+\mathrm{35}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{49}}\int\:\:\frac{\mathrm{2}{x}−\mathrm{9}}{{x}}\:{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{49}}{x}−\frac{\mathrm{9}}{\mathrm{49}}{ln}\mid{x}\mid\:+{C} \\ $$$$=\frac{\mathrm{2}}{\mathrm{49}}\left(\mathrm{7}{t}+\mathrm{22}\right)−\frac{\mathrm{9}}{\mathrm{49}}{ln}\mid\mathrm{7}{t}+\mathrm{22}\mid\:+{C} \\ $$

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