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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int(\text{x}-2)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$
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Answer

Let $\text{I}=\int(\text{x}-2)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$Also, $\text{x}-2=\lambda\frac{\text{d}}{\text{dx}}(2\text{x}^2-6\text{x}+5)+\mu$
$=(4\lambda)\text{x}+\mu-6\lambda$ Equating the co-efficient of like terms $4\lambda=1\Rightarrow\lambda=\frac{1}{4}$And
$\mu-6\lambda=-2$ $\Rightarrow\mu-6\times\frac{1}{4}=-2$ $\Rightarrow\mu=-2+\frac{3}{2}=-\frac{1}{2}$ $\therefore\ \text{I}=\int\Big[\frac{1}{4}(4\text{x}-6)-\frac{1}{2}\Big]\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$ $=\frac{1}{4}\int(4\text{x}-6)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}\\-\frac{1}{2}\int\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$ Let $2\text{x}^2-6\text{x}+5=\text{t}$ $\Rightarrow(4\text{x}-6)\text{dx = dt}$ $\therefore\ \text{I}=\frac{1}{4}\int\text{t}^{\frac{1}{2}}\text{dt}-\frac{1}{2}\int\sqrt{2\Big(\text{x}^2-3\text{x}+\frac{5}{2}\Big)}\text{dx}$ $=\frac{1}{4}\int\text{t}^{\frac{1}{2}}-\frac{\sqrt2}{2}\int\sqrt{\text{x}^2-3\text{x}\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2+\frac{5}{2}}\text{dx}$ $=\frac{1}{4}\Bigg[\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}\Bigg]-\frac{1}{\sqrt2}\int\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2-\frac{9}{4}+\frac{5}{2}}\text{dx}$ $=\frac{1}{6}\text{t}^{\frac{3}{2}}-\frac{1}{\sqrt2}\int\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2-\frac{9+10}{4}}\text{dx}$ $=\frac{1}{6}\text{t}^{\frac{3}{2}}-\frac{1}{\sqrt2}\int\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}\text{dx}$ $=\frac{1}{6}\big(2\text{x}^2-6\text{x}+5)^{\frac{3}{2}}-\frac{1}{\sqrt2}\bigg[\bigg(\frac{\text{x}-\frac{3}{2}}{2}\bigg)\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}\\+\frac{1}{8}\log\bigg|\Big(\text{x}-\frac{3}{2}\Big)+\sqrt{\text{x}^2-3\text{x}+\frac{5}{2}}\bigg|\bigg]+\text{C}$ $=\frac{1}{6}\big(2\text{x}^2-6\text{x}+5)^{\frac{3}{2}}-\frac{1}{\sqrt2}\bigg[\frac{2\text{x}-3}{4}\sqrt{\text{x}^2-3\text{x}+\frac{5}{2}}\\+\frac{1}{8}\log\bigg|\frac{2\text{x}-3}{2}+\sqrt{\text{x}^2-3\text{x}+\frac{5}{2}}\bigg|\bigg]+\text{C}$

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