question_answer

Evaluate \[\int{\frac{({{x}^{2}}+1)({{x}^{2}}+2)}{({{x}^{2}}+3)({{x}^{2}}+4)}dx.}\]

Answer:

Let \[l=\int{\frac{({{x}^{2}}+1)({{x}^{2}}+2)}{({{x}^{2}}+3)({{x}^{2}}+4)}}dx\] Here, the degree of numerator and denominator is 4. So, we convert the integral into simple form by putting \[{{x}^{2}}=t.\] \[\therefore \]            \[\frac{({{x}^{2}}+1)({{x}^{2}}+2)}{({{x}^{2}}+3)({{x}^{2}}+4)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}\]             \[=\frac{{{t}^{2}}+3t+2}{{{t}^{2}}+7t+12}\] Since, degree of numerator and denominator is same, so it can be written as             \[=\frac{{{t}^{2}}+3t+2}{{{t}^{2}}+7t+12}=1-\frac{4t+10}{{{t}^{2}}+7t+12}\]              [on dividing numerator by denominator] Let        \[\frac{4t+10}{(t+4)(t+3)}=\frac{A}{(t+4)}+\frac{B}{(t+3)}\] \[\Rightarrow \]   \[\frac{4t+10}{(t+4)(t+3)}=\frac{A(t+3)+B(t+4)}{(t+4)(t+3)}\] \[\Rightarrow \]   \[4t+10=At+3A+Bt+4B\] \[\Rightarrow \]   \[4t+10=t(A+B)+(3A+4B)\] On comparing the coefficients of t and constant terms from both sides, we get             \[A+B=4\]                                ?(i) and       \[3A+4B=10\]                           ?(ii) On multiplying Eq. (i) by 3 and then subtracting Eq. (ii) from Eq. (i), we get  \[-\,B=2\,\,\,\,\Rightarrow \,\,\,\,B=-\,2\] Then, from Eq. (i), we get A = 6 \[\Rightarrow \]   \[\frac{4t+10}{(t+4)(t+3)}=\frac{6}{(t+4)}-\frac{2}{(t+3)}\] \[\therefore \] \[l=\int{1\,dx-\left[ \int{\left( \frac{6}{{{x}^{2}}+4}-\frac{2}{{{x}^{2}}+3} \right)\,dx} \right]}\]    \[[\text{put}\,\,t={{x}^{2}}]\] \[=x-6\int{\frac{1}{{{x}^{2}}+{{(2)}^{2}}}dx+2\int{\frac{1}{{{x}^{2}}+{{(\sqrt{3})}^{2}}}}}dx\] \[=x-6\times \frac{1}{2}{{\tan }^{-1}}\frac{x}{2}+2\times \frac{1}{\sqrt{3}}{{\tan }^{-1}}\frac{x}{\sqrt{3}}+C\] \[\left[ \because \,\,\,\int{\frac{1}{{{a}^{2}}+{{x}^{2}}}\,dx=\frac{1}{a}{{\tan }^{-1}}\frac{x}{a}+C} \right]\]\[=x-3{{\tan }^{-1}}\frac{x}{2}+\frac{2}{\sqrt{3}}{{\tan }^{-1}}\frac{x}{\sqrt{3}}+C\]