1 4 x^2-20 x+17 d x - Vidyadip

Question

$\int \frac{1}{4 x^2-20 x+17} d x$

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Answer

$\text { Let } I =\int \frac{1}{4 x^2-20 x+17} d x$
$=\int \frac{1}{4\left(x^2-5 x+\frac{17}{4}\right)} d x$
$\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left(\frac{1}{2} \times(-5)\right)^2$
$=\frac{25}{4}$
$\therefore I=\frac{1}{4} \int \frac{1}{x^2-5 x+\frac{25}{4}-\frac{25}{4}+\frac{17}{4}} d x$
$=\frac{1}{4} \int \frac{1}{\left(x-\frac{5}{2}\right)^2-2} d x$
$=\frac{1}{4} \int \frac{1}{\left(x-\frac{5}{2}\right)^2-(\sqrt{2})^2} d x$
$=\frac{1}{4} \cdot \frac{1}{2 \sqrt{2}} \log \left|\frac{x-\frac{5}{2}-\sqrt{2}}{x-\frac{5}{2}+\sqrt{2}}\right|+ c$
$\therefore I=\frac{1}{8 \sqrt{2}} \log \left|\frac{2 x-5-2 \sqrt{2}}{2 x-5+2 \sqrt{2}}\right|+ c$

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