Q113 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

If $\int e^x\left(\frac{x^2-8 x+19}{(x-1)^5}\right) d x=\frac{e^x(l x+m)}{(x-1)^4}+C$, then $4 l+m=$

  • A
    -5
  • B
    -2
  • C
    1
  • Correct answer
    D
    0

Solution

We have $\begin{aligned} & \int e^x\left(\frac{x^2-8 x+19}{(x-1)^5}\right) d x=\frac{e^x(l x+m)}{(x-1)^4}+C \\ & \text { Let } I=\int e^x\left(\frac{x^2-5 x+4-3 x+15}{(x-1)^5}\right) d x \\ & I=\int e^x\left(\frac{(x-4)(x-1)}{(x-1)^5}-3 \frac{(x-5)}{(x-1)^5}\right) d x \\ & I=\int e^x\left(\frac{x-4}{(x-1)^4}-\frac{3(x-5)}{(x-1)^5}\right) d x \\ & \end{aligned}$ $\begin{aligned} & \text { Here, } f(x)=\frac{x-4}{(x-1)^4} \\ & \Rightarrow \quad f(x)=\frac{-3(x-5)}{(x-1)^5} \\ & \therefore \quad I=\frac{e^x(x-4)}{(x-1)^4}+C \\ & \therefore \quad l=1 \quad m=-4 \\ & \therefore \quad 4 l+m=4-4=0 \\ & \end{aligned}$

Hence, the correct answer is option: 'D'.

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