Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{5x+7}{\sqrt{\left(x-5\right)\left(x-4\right)}}\right)dx \\ =\int \left(\frac{5x+7}{\sqrt{x^2-9x+20}}\right)dx \\ \mathrm{Let}5x+7=A\frac{d}{dx}\left(x^2-9x+20\right)+B \\ \Rightarrow 5x+7=A\left(2x-9\right)+B \\ \mathrm{Equating}\mathrm{Coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2A=5 \\ \Rightarrow A=\frac{5}{2} \\ \mathrm{And} \\ -9A+B=7 \\ \Rightarrow -9\times \frac{5}{2}+B=7 \\ \Rightarrow B=7+\frac{45}{2} \\ \Rightarrow B=\frac{59}{2} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{5}{2}\left(2x-9\right)+\frac{59}{2}}}{\sqrt{x^2-9x+20}}\right)dx \\ =\frac{5}{2}\int \frac{\left(2x-9\right)dx}{\sqrt{x^2-9x+20}}+\frac{59}{2}\int \frac{dx}{\sqrt{x^2-9x+20}} \\ \mathrm{Putting}{x}^2-9x+20=t \\ \Rightarrow \left(2x-9\right)dx=dt \\ I=\frac{5}{2}\int \frac{dt}{\sqrt{t}}+\frac{59}{2}\int \frac{dx}{\sqrt{x^2-9x+{\left({\displaystyle \frac{9}{2}}\right)}^2-{\left({\displaystyle \frac{9}{2}}\right)}^2+20}} \\ =\frac{5}{2}\int t^{-\frac{1}{2}}dt+\frac{59}{2}\int \frac{dx}{\sqrt{{\left(x-{\displaystyle \frac{9}{2}}\right)}^2-{\displaystyle \frac{81+80}{4}}}} \\ =\frac{5}{2}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+\frac{59}{2}\int \frac{dx}{\sqrt{{\left(x-{\displaystyle \frac{9}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2}} \\ =\frac{5}{2}\times 2\sqrt{t}+\frac{59}{2}\log \left|\left(x-\frac{9}{2}\right)+\sqrt{{\left(x-\frac{9}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}\right|+C \\ =5\sqrt{t}+\frac{59}{2}\log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9x+20}\right|+C \\ =5\sqrt{x^2-9x+20}+\frac{59}{2}\log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9x+20}\right|+C \end{gathered}$$