Question:
Integrate the function: \(\frac{(x-3)e^x}{(x-1)^3}\)
Integrate the function: \(\frac{(x-3)e^x}{(x-1)^3}\)
Updated On: Oct 4, 2023
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Solution and Explanation
The correct answer is: \(∫e^x[\frac{(x-3)}{(x-1)^2}]dx=\frac{e^x}{(x-1)^2}+C\)
\(∫e^x=[\frac{x-3}{(x-1)^3}]dx=∫e^x[\frac{x-1-2}{(x-1)^3}]dx\)
\(=∫e^x[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}]dx\)
Let \(ƒ(x)=\frac{1}{(x-1)^2}\,\,ƒ'(x)=\frac{-2}{(x-1)^3}\)
It is known that,\(∫e^x[ƒ(x)+ƒ'(x)]dx=e^x ƒ(x)+C\)
\(∴∫e^x[\frac{(x-3)}{(x-1)^2}]dx=\frac{e^x}{(x-1)^2}+C\)
\(∫e^x=[\frac{x-3}{(x-1)^3}]dx=∫e^x[\frac{x-1-2}{(x-1)^3}]dx\)
\(=∫e^x[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}]dx\)
Let \(ƒ(x)=\frac{1}{(x-1)^2}\,\,ƒ'(x)=\frac{-2}{(x-1)^3}\)
It is known that,\(∫e^x[ƒ(x)+ƒ'(x)]dx=e^x ƒ(x)+C\)
\(∴∫e^x[\frac{(x-3)}{(x-1)^2}]dx=\frac{e^x}{(x-1)^2}+C\)
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