Question

Let $f\left(x\right)=\int \frac{e^x}{x}dx$ and $\int \frac{\left(e^{x-1}\right)\left(2x\right)}{x^2-5x+4}dx=\alpha f(x-4)+\beta f(x-1)+\gamma$, then

• A. $\ln 3\alpha =3$
• B. $4+3\beta =\ln 3\alpha$
• C. $3\beta +2=0$
• D. $\ln 3\alpha =3+\ln 8$

Answer 1

Answer: (C), (D)

The correct options are
C $\ln 3\alpha =3+\ln 8$
D $3\beta +2=0$

$\int \frac{e^{x-1}.\left(2x\right)}{x^2-5x+4}dx$
$=\int \frac{e^{x-1}[(x-1)+(x-4)+5]}{(x-1)(x-4)}dx=\int \frac{e^{x-1}}{x-4}dx+\int \frac{e^{x-1}}{x-1}dx+5\int \frac{e^{x-1}}{(x-1)(x-4)}dx=e^3\int \frac{e^{x-4}}{x-4}dx+\int \frac{e^{x-1}}{x-1}dx+\frac{5}{3}\int \frac{e^{x-1}[(x-1)-(x-4)]}{(x-1)(x-4)}dx=e^3\int \frac{e^{x-4}}{x-4}dx+\int \frac{e^{x-1}}{x-1}dx+\frac{5}{3}{e}^3\int \frac{e^{x-4}}{x-4}dx-\frac{5}{3}\int \frac{e^{x-1}}{x-1}dx=\frac{8}{3}{e}^3\int \frac{e^{x-4}}{x-4}dx-\frac{2}{3}\int \frac{e^{x-1}}{x-1}dx=\frac{8}{3}{e}^{3}f(x-4)+(-\frac{2}{3})f(x-1)+\gamma \dots (\because f\left(x\right)=\int \frac{e^x}{x}dx)$
Comparing it with the given equation $\int \frac{\left(e^{x-1}\right)\left(2x\right)}{x^2-5x+4}dx=\alpha f(x-4)+\beta f(x-1)+\gamma$, we get
$3\alpha =8e^3\Rightarrow \ln 3\alpha =\ln 8+3$
$3\beta =-2\Rightarrow 3\beta +2=0$