Question

\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]

Answer 1

 

`   \text{ Let I } = ∫   { x   dx}/{\sqrt{x^2 + x + 1}}`
\[\text{ Consider, } \]
\[x = A \frac{d}{dx} \left( x^2 + x + 1 \right) + B\]
\[ \Rightarrow x = A \left( 2x + 1 \right) + B\]
\[ \Rightarrow x = \left( 2A \right) x + A + B\]
\[\text{Equating Coefficient of like terms}\]
\[2A = 1\]
\[ \Rightarrow A = \frac{1}{2}\]
\[\text{ And }\]
\[A + B = 0\]
\[ \Rightarrow \frac{1}{2} + B = 0\]
\[ \Rightarrow B = - \frac{1}{2}\]
\[ \therefore I = \int\frac{\left( \frac{1}{2} \left( 2x + 1 \right) - \frac{1}{2} \right)}{\sqrt{x^2 + x + 1}} dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 1}{\sqrt{x^2 + x + 1}} \right)dx - \frac{1}{2}\int\frac{dx}{\sqrt{x^2 + x + \frac{1}{4} - \frac{1}{4} + 1}}\]
\[\text{ Putting x }^2 + x + 1 = t\]
\[ \Rightarrow \left( 2x + 1 \right) dx = dt\]
\[\text{ Then, } \]
\[I = \frac{1}{2}\int\frac{dt}{\sqrt{t}} - \frac{1}{2}\int\frac{dx}{\sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}\]
\[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt - \frac{1}{2} \text{ log  }\left| x + \frac{1}{2} + \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| + C\]
\[ = \frac{1}{2}\left| \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right| - \frac{1}{2} \text{ log }\left| x + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C\]
\[ = \sqrt{t} - \frac{1}{2} \text{ log  }\left| x + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C\]
\[ = \sqrt{x^2 + x + 1} - \frac{1}{2} \text{ log }\left| x + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C\]