Question

Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .

Answer 1

\[I = \int\left( 3x - 2 \right)\sqrt{x^2 + x + 1}dx\]

\[\text { Let }3x - 2 = a(2x + 1) + b\]

\[ \Rightarrow a = \frac{3}{2} \text { and } b = \frac{- 7}{2}\]

\[\text { So,} I = \frac{3}{2}\int\left( 2x + 1 \right)\sqrt{x^2 + x + 1}dx - \frac{7}{2}\int\sqrt{x^2 + x + 1}dx\]

\[\text { Let } I = \frac{3}{2} I_1 - \frac{7}{2} I_2 \ldots\left( 1 \right)\]

\[\text{ Here }, I_1 = \int\left( 2x + 1 \right)\sqrt{x^2 + x + 1}dx \text { and } I_2 = \int\sqrt{x^2 + x + 1}dx\]

\[\text { Now }, I_1 = \int\left( 2x + 1 \right)\sqrt{x^2 + x + 1}dx\]

\[ \text { Let } x^2 + x + 1 = t\]

\[ \Rightarrow \left( 2x + 1 \right)dx = dt\]

\[\text { So }, I_1 = \int\sqrt{t} dt = \frac{2}{3} t^\frac{3}{2} = \frac{2}{3}( x^2 + x + 1 )^\frac{3}{2} + c_1 \ldots\left( 2 \right)\]

\[\text { And } I_2 = \int\sqrt{x^2 + x + 1}dx = \int\sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dx \]

\[ = \frac{x + \frac{1}{2}}{2}\sqrt{x^2 + x + 1} + \frac{3}{8}\log\left| x + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + c_2 \ldots\left( 3 \right)\]

By putting the values of equation (2) and equation (3) in equation (1), we get:

\[I = \left( x^2 + x + 1 \right)^\frac{3}{2} - \frac{7}{2}\left[ \left( \frac{2x + 1}{4} \right)\sqrt{x^2 + x + 1} + \frac{3}{8}\log\left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x + 1} \right| \right] + c\]