Evaluate the following integrals: (2x-5) x^2-4x+3 dx - Vidyadip

Question

Evaluate the following integrals:
$\int(2\text{x}-5)\sqrt{\text{x}^2-4\text{x}+3}\text{dx}$

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Answer

Let $\text{I}=\int(2\text{x}-5)\sqrt{\text{x}^2-4\text{x}+3}\text{dx}$ $=\int(2\text{x}-4-1)\sqrt{\text{x}^2-4\text{x}+3}\text{dx}$$=\int(2\text{x}+4)\sqrt{\text{x}^2-4\text{x}+3}\text{dx}-\int\sqrt{\text{x}^2-4\text{x}+3}\text{dx}$
$=(2\text{x}-4)\sqrt{\text{x}^2-4\text{x}+3}\text{dx}-\int\sqrt{\text{x}^2-4\text{x}+4-4+3}\text{dx}$
$=\int(2\text{x}-4)\sqrt{\text{x}^2-4\text{x}+3}\text{dx}-\int\sqrt{(\text{x}-2)^2-1^2}\text{dx}$
Let $\text{x}^2-4\text{x}+3=\text{t}$
$\Rightarrow({2\text{x}-4})\text{dx = dt}$
$\therefore\ \text{I}=\int\sqrt{\text{t}}\text{dt}-\int\sqrt{(\text{x}-2)^2-1^2}\text{dx}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}-\Big[\frac{\text{x}-2}{2}\sqrt{(\text{x}-2)^2-1^2}-\frac{1^2}{2}\log\Big|(\text{x}-2)\\+\sqrt{(\text{x}-2)^2-1}\Big|\Big]+\text{C}$
$=\frac{2}{3}(\text{x}^2-4\text{x}+3)^{\frac{3}{2}}-\Big(\frac{\text{x}-2}{2}\Big)\sqrt{\text{x}^2-4\text{x}+3}\\+\frac{1}{2}\log\Big|(\text{x}-2)+\sqrt{\text{x}^2-4\text{x}+3}\Big|+\text{C}$

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