question_answer

Evaluate \[\int{\frac{1}{3{{x}^{2}}+5x+7}\,dx.}\]

OR

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

Answer:

Let \[l=\int{\frac{1}{3{{x}^{2}}+\,5x+7}}\,dx=\int{\frac{dx}{3\left( {{x}^{2}}+\frac{5x}{3}+\frac{7}{3} \right)}}\]

\[=\frac{1}{3}\int{\frac{dx}{{{x}^{2}}+\frac{5x}{3}+\frac{7}{3}}}\]

\[=\frac{1}{3}\int{\frac{dx}{{{x}^{2}}+\frac{5x}{3}+\frac{7}{3}+\frac{25}{36}-\frac{25}{36}}}\]

\[=\frac{1}{3}\int{\frac{dx}{{{\left( x+\frac{5}{6} \right)}^{2}}+\left( \frac{7}{3}-\frac{25}{36} \right)}}\]

\[=\frac{1}{3}\int{\frac{dx}{{{\left( x+\frac{5}{6} \right)}^{2}}+\left( \frac{84-25}{36} \right)}}\]

\[=\frac{1}{3}\int{\frac{dx}{{{\left( x+\frac{5}{6} \right)}^{2}}+{{\left( \frac{\sqrt{59}}{36} \right)}^{2}}}}\]

\[=\frac{1}{3}\cdot \frac{6}{\sqrt{59}}{{\tan }^{-1}}\left( \frac{x+\frac{5}{6}}{\frac{\sqrt{59}}{6}} \right)+C\]

\[\left[ \because \,\,\int{\frac{dx}{{{a}^{2}}+{{x}^{2}}}=\frac{1}{a}{{\tan }^{-1}}\frac{x}{a}} \right]\]

\[\therefore \]      \[l=\frac{2}{\sqrt{59}}{{\tan }^{-1}}\left( \frac{6x+5}{\sqrt{59}} \right)+C\]

OR

Let \[l=\int{\frac{x{{e}^{x}}}{{{(x\,+1)}^{2}}}}\,dx\]

\[=\int{\frac{(x+1-1)\,{{e}^{x}}}{{{(x\,+1)}^{2}}}}\,dx\]

\[=\int{\left[ \frac{x+1}{{{(x\,+1)}^{2}}}-\frac{1}{{{(x\,+1)}^{2}}} \right]}\,{{e}^{x}}dx\]

\[=\int{\left[ \frac{1}{x+1}-\frac{1}{{{(x\,+1)}^{2}}} \right]}\,{{e}^{x}}dx\]

Now, consider \[f(x)=\frac{1}{1+x},\] then \[f'(x)=\frac{-\,1}{{{(1+x)}^{2}}}.\] Thus, the given integrand is of the form

\[\int{{{e}^{x}}(f(x)+f'(x))\,dx.}\]

Hence, \[l=\frac{{{e}^{x}}}{x+1}+C\]

\[[\because \,\,{{e}^{x}}\{f(x)+f'(x)\}dx={{e}^{x}}(x)]\]