Question Number 492 by 123456 last updated on 25/Jan/15
$$\int\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}{dx} \\ $$
Answered by prakash jain last updated on 14/Jan/15
![x=t^2 dx=2t dt ∫((1−t)/(1+t))∙2t dt=2∫ ((t−t^2 )/(1+t)) dt =2∫((1+t+1−t^2 −2)/(1+t)) =2[∫1 dt+∫(1−t)dt−∫(2/(1+t)) dt] =2[t+t−(t^2 /2)−2ln ∣1+t∣]+C =4t−t^2 −4ln ∣1+t∣+C =4(√x)−x−4ln ∣1+(√x)∣+C](https://www.tinkutara.com/question/Q493.png)
$${x}={t}^{\mathrm{2}} \\ $$$${dx}=\mathrm{2}{t}\:{dt} \\ $$$$\int\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\centerdot\mathrm{2}{t}\:{dt}=\mathrm{2}\int\:\frac{{t}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}}\:{dt} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}+{t}+\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{2}}{\mathrm{1}+{t}} \\ $$$$=\mathrm{2}\left[\int\mathrm{1}\:{dt}+\int\left(\mathrm{1}−{t}\right){dt}−\int\frac{\mathrm{2}}{\mathrm{1}+{t}}\:{dt}\right] \\ $$$$=\mathrm{2}\left[{t}+{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2ln}\:\mid\mathrm{1}+{t}\mid\right]+{C} \\ $$$$=\mathrm{4}{t}−{t}^{\mathrm{2}} −\mathrm{4ln}\:\mid\mathrm{1}+{t}\mid+{C} \\ $$$$=\mathrm{4}\sqrt{{x}}−{x}−\mathrm{4ln}\:\mid\mathrm{1}+\sqrt{{x}}\mid+{C} \\ $$