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Question

Mathematics Question on integral

Integrate the function: (x−3)ex(x−1)3\frac{(x-3)e^x}{(x-1)^3}

Answer

The correct answer is: ∫ex[(x−3)(x−1)2]dx=ex(x−1)2+C∫e^x[\frac{(x-3)}{(x-1)^2}]dx=\frac{e^x}{(x-1)^2}+C
∫ex=[x−3(x−1)3]dx=∫ex[x−1−2(x−1)3]dx∫e^x=[\frac{x-3}{(x-1)^3}]dx=∫e^x[\frac{x-1-2}{(x-1)^3}]dx
=∫ex[1(x−1)2−2(x−1)3]dx=∫e^x[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}]dx
Let ƒ(x)=1(x−1)2  ƒ′(x)=−2(x−1)3ƒ(x)=\frac{1}{(x-1)^2}\,\,ƒ'(x)=\frac{-2}{(x-1)^3}
It is known that,∫ex[ƒ(x)+ƒ′(x)]dx=exƒ(x)+C∫e^x[ƒ(x)+ƒ'(x)]dx=e^x ƒ(x)+C
∴∫ex[(x−3)(x−1)2]dx=ex(x−1)2+C∴∫e^x[\frac{(x-3)}{(x-1)^2}]dx=\frac{e^x}{(x-1)^2}+C