Question

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

Answer 1

\[\text{We have}, \]

\[I = \int\left( \frac{5x + 7}{\sqrt{\left( x - 5 \right)\left( x - 4 \right)}} \right) dx\]

\[ = \int\left( \frac{5x + 7}{\sqrt{x^2 - 9x + 20}} \right) dx\]

\[\text{ Let  5x + 7 }= A \frac{d}{dx} \left( x^2 - 9x + 20 \right) + B\]

\[ \Rightarrow 5x + 7 = A \left( 2x - 9 \right) + B\]

\[\text{Equating Coefficients of like terms}\]

\[2A = 5\]

\[ \Rightarrow A = \frac{5}{2}\]

\[\text{ And }\]

\[ - 9A + B = 7\]

\[ \Rightarrow - 9 \times \frac{5}{2} + B = 7\]

\[ \Rightarrow B = 7 + \frac{45}{2}\]

\[ \Rightarrow B = \frac{59}{2}\]

\[ \therefore I = \int\left( \frac{\frac{5}{2} \left( 2x - 9 \right) + \frac{59}{2}}{\sqrt{x^2 - 9x + 20}} \right) dx\]

\[ = \frac{5}{2}\int\frac{\left( 2x - 9 \right) dx}{\sqrt{x^2 - 9x + 20}} + \frac{59}{2}\int\frac{dx}{\sqrt{x^2 - 9x + 20}}\]

\[\text{ Putting x}^2 - 9x + 20 = t\]

\[ \Rightarrow \left( 2x - 9 \right) dx = dt\]

\[I = \frac{5}{2}\int\frac{dt}{\sqrt{t}} + \frac{59}{2}\int\frac{dx}{\sqrt{x^2 - 9x + \left( \frac{9}{2} \right)^2 - \left( \frac{9}{2} \right)^2 + 20}}\]

\[ = \frac{5}{2}\int t^{- \frac{1}{2}} \text{ dt }+ \frac{59}{2}\int\frac{dx}{\sqrt{\left( x - \frac{9}{2} \right)^2 - \frac{81 + 80}{4}}}\]

\[ = \frac{5}{2} \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \frac{59}{2} \int\frac{dx}{\sqrt{\left( x - \frac{9}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}\]

\[ = \frac{5}{2} \times 2\sqrt{t} + \frac{59}{2} \text{ log }\left| \left( x - \frac{9}{2} \right) + \sqrt{\left( x - \frac{9}{2} \right)^2 - \left( \frac{1}{2} \right)^2} \right| + C\]

\[ = 5\sqrt{t} + \frac{59}{2} \text{ log} \left| \left( x - \frac{9}{2} \right) + \sqrt{x^2 - 9x + 20} \right| + C\]

\[ = 5\sqrt{x^2 - 9x + 20} + \frac{59}{2} \text{ log }\left| \left( x - \frac{9}{2} \right) + \sqrt{x^2 - 9x + 20} \right| + C\]