Question Number 171038 by Mastermind last updated on 06/Jun/22
Answered by MJS_new last updated on 07/Jun/22
![∫_4 ^9 ((x+1)/(x+2(√x)−3))dx=∫_4 ^9 ((x+1)/(((√x)+1)^2 −4))dx= [t=(√x)+1 → dx=2(√x)dt] =2∫_3 ^4 (((t−1)(t^2 −2t+2))/(t^2 −4))dt= =∫_3 ^4 (2t−6+((15)/(t+2))+(1/(t+2)))dt= =[t^2 −6t+15ln ∣t+2∣ +ln ∣t−2∣]_3 ^4 = =1+16ln 2 +15ln 3 −15ln 5](Q171061.png)
$$\underset{\mathrm{4}} {\overset{\mathrm{9}} {\int}}\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}\sqrt{{x}}−\mathrm{3}}{dx}=\underset{\mathrm{4}} {\overset{\mathrm{9}} {\int}}\frac{{x}+\mathrm{1}}{\left(\sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\mathrm{1}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}\frac{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}\right)}{{t}^{\mathrm{2}} −\mathrm{4}}{dt}= \\ $$$$=\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}\left(\mathrm{2}{t}−\mathrm{6}+\frac{\mathrm{15}}{{t}+\mathrm{2}}+\frac{\mathrm{1}}{{t}+\mathrm{2}}\right){dt}= \\ $$$$=\left[{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{15ln}\:\mid{t}+\mathrm{2}\mid\:+\mathrm{ln}\:\mid{t}−\mathrm{2}\mid\right]_{\mathrm{3}} ^{\mathrm{4}} = \\ $$$$=\mathrm{1}+\mathrm{16ln}\:\mathrm{2}\:+\mathrm{15ln}\:\mathrm{3}\:−\mathrm{15ln}\:\mathrm{5} \\ $$
Commented by peter frank last updated on 07/Jun/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by GalaxyBills last updated on 07/Jun/22
