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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration3 Marks
Question
Evaluate : $\int \frac{1}{\sqrt{3 x^2-4 x+2}} \cdot d x$
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Answer

$=\int \frac{1}{\sqrt{3\left(x^2-\frac{4}{3} x+\frac{2}{3}\right)}} \cdot d x$ $\because\left\{\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left(\frac{1}{2}\left(-\frac{4}{3}\right)\right)^2=\left(-\frac{2}{3}\right)^2=\frac{4}{9}\right\}$
$\begin{aligned} & =\int \frac{1}{\sqrt{3} \cdot \sqrt{x^2-\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{2}{3}}} \cdot d x \\ & =\frac{1}{\sqrt{3}} \cdot \int \frac{1}{\sqrt{\left(x^2-\frac{4}{3} x+\frac{4}{9}\right)+\left(\frac{2}{3}-\frac{4}{9}\right)}} \cdot d x \\ & =\frac{1}{\sqrt{3}} \cdot \int \frac{1}{\sqrt{\left(x-\frac{2}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2} \cdot d x}\end{aligned}$
$\begin{aligned} & \because \int \frac{1}{\sqrt{x^2+a^2}} \cdot d x=\log \left|x+\sqrt{x^2+a^2}\right|+c \\ & =\frac{1}{\sqrt{3}} \cdot \log \left(\left(x-\frac{2}{3}\right)+\sqrt{\left(x-\frac{2}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2}\right)+c \\ & =\frac{1}{\sqrt{3}} \cdot \log \left(\left(x-\frac{2}{3}\right)+\sqrt{x^2-\frac{4}{3} x+\frac{2}{3}}\right)+c\end{aligned}$

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