question_answer

Evaluate \[\int{(x\,+1)}\sqrt{1-x-{{x}^{2}}}dx\]

Answer:

Let \[l=\int{(x\,+1)}\sqrt{1-x-{{x}^{2}}}dx\] Again, let \[x+1=\lambda \frac{d}{dx}(1-x-{{x}^{2}})+\mu \] \[\Rightarrow \]   \[x+1=\lambda (-1-2x)+\mu \] On comparing the coefficient of x and the constant term from both sides, we get             \[-\,2\lambda =1\] and \[\mu -\lambda =1\] \[\Rightarrow \]   \[\lambda =\frac{-\,1}{2}\] and \[\mu =\frac{1}{2}\] \[\therefore \]      \[l=\int{\left[ -\frac{1}{2}(-\,1-\,2x)+\frac{1}{2} \right]}\sqrt{1-x-{{x}^{2}}}dx\] \[=-\frac{1}{2}\int{(-\,1-\,2x)\sqrt{1-x-{{x}^{2}}}dx}\]                         \[+\frac{1}{2}\int{\sqrt{1-x-{{x}^{2}}}dx}\] \[=-\frac{1}{2}\int{(-\,1-\,2x)\sqrt{1-x-{{x}^{2}}}dx}\] \[+\frac{1}{2}\int{\sqrt{\left\{ 1-\left( {{x}^{2}}+x+\frac{1}{4}-\frac{1}{4} \right) \right\}}dx}\]\[=-\frac{1}{2}\int{\sqrt{t}}dt+\frac{1}{2}\int{\sqrt{{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}-{{\left( x+\frac{1}{2} \right)}^{2}}}dx}\]                                     [where \[t=1-x-{{x}^{2}}\]] \[=-\frac{1}{2}\left[ \frac{{{t}^{3/2}}}{3/2} \right]+\frac{1}{2}\left\{ \frac{1}{2}\left( x+\frac{1}{2} \right)\sqrt{1-x-{{x}^{2}}} \right.\]                 \[\left. +\frac{1}{2}\times \frac{5}{4}{{\sin }^{-1}}\left( \frac{x+1/2}{\sqrt{5}/2} \right) \right\}+C\] \[=-\frac{1}{3}{{(1-x-{{x}^{2}})}^{3/2}}+\frac{1}{8}(2x+1)\sqrt{1-x-{{x}^{2}}}\] \[+\frac{5}{16}{{\sin }^{-1}}\left( \frac{2x+1}{\sqrt{5}} \right)+C\]