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dx-x-x-




Question Number 20243 by tammi last updated on 24/Aug/17
∫(dx/(x+(√x)))
$$\int\frac{{dx}}{{x}+\sqrt{{x}}} \\ $$
Answered by $@ty@m last updated on 25/Aug/17
Here I=∫((1/( (√x)))/( (√x)+1))dx (dividing N^r  & D^r  by (√x))  Let (√x)+1=t  ((−1)/(2(√x)))dx=dt  ∴ I=∫((−2dt)/t)  =−2ln∣t∣+C  =−2ln∣(√x)+1∣+C
$${Here}\:{I}=\int\frac{\frac{\mathrm{1}}{\:\sqrt{{x}}}}{\:\sqrt{{x}}+\mathrm{1}}{dx}\:\left({dividing}\:{N}^{{r}} \:\&\:{D}^{{r}} \:{by}\:\sqrt{{x}}\right) \\ $$$${Let}\:\sqrt{{x}}+\mathrm{1}={t} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}={dt} \\ $$$$\therefore\:{I}=\int\frac{−\mathrm{2}{dt}}{{t}} \\ $$$$=−\mathrm{2}{ln}\mid{t}\mid+{C} \\ $$$$=−\mathrm{2}{ln}\mid\sqrt{{x}}+\mathrm{1}\mid+{C} \\ $$

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