Question

Find : $\int \frac{(2x-5)e^{2x}}{(2x-3{)}^3}dx.$ OR Find : $\int \frac{(x^2+x+1)}{(x^2+1)(x+2)}dx.$

Answer 1

Let $I=\int \frac{(2x-5)e^{2x}}{(2x-3{)}^3}dxPut2x=t\Rightarrow dx=\frac{1}{2}dt$

$\therefore I=\frac{1}{2}\int \frac{(t-5)e^t}{(t-3{)}^3}dt\Rightarrow 2I=\int \left[\begin{array}{c}\frac{(t-3)-2}{(t-3{)}^3}\end{array}\right]e^{t}dt\Rightarrow 2I=\int \left[\begin{array}{c}\frac{1}{(t-3{)}^2}+\frac{-2}{(t-3{)}^3}\end{array}\right]e^{t}dt\Rightarrow 2I=\int \frac{1}{(t-3{)}^2}\times e^{t}dt-\int \left[\begin{array}{c}\frac{2}{(t-3{)}^3}\end{array}\right]e^{t}dt\Rightarrow 2I=\frac{1}{(t-3{)}^2}\times \int e^{t}dt-\int \left(\begin{array}{c}\frac{d}{dt}\frac{1}{(t-3{)}^2}\int e^{t}dt\end{array}\right)dt-\int \left[\begin{array}{c}\frac{2}{(t-3{)}^3}\end{array}\right]e^{t}dt\Rightarrow 2I=\frac{1}{(t-3{)}^2}\times e^t-\int \left(\begin{array}{c}-\frac{2}{(t-3{)}^2}\times e^t\end{array}\right)dt-\int \left[\begin{array}{c}\frac{2}{(t-3{)}^3}\end{array}\right]e^{t}dt=\frac{1}{(t-3{)}^2}\times e^t+C\therefore I=\frac{1}{2(2x-3{)}^2}\times e^{2x}+C.$

OR

$I=\int \frac{x^2+x+1}{(x^2+1)(x+2)}dxConsider\frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{Ax+B}{x^2+1}+\frac{C}{x+2}\Rightarrow x^2+x+1=(Ax+B)(x+2)+C(x^2+1)Oncomparingliketerms,weget:A=\frac{2}{5},B=\frac{1}{5},C=\frac{3}{5}\therefore I=\frac{1}{5}\int \frac{2x}{(x^2+1)}dx+\frac{1}{5}\int \frac{1}{(x^2+1)}dx+\frac{3}{5}\int \frac{1}{(x+2)}dx\Rightarrow 1=\frac{1}{5}log|x^2+1|+\frac{1}{5}ta{n}^{-1}x+\frac{3}{5}log|x+2|.$