Mathematics

# If $\displaystyle\int \dfrac{(x-1)dx}{x^{2}\sqrt{2x^{2}-2x+1}}=\dfrac{\sqrt{f(x)}}{g(x)}+C$, where $f(x)$ is a quadratic expression and $g(x)$ is a monic linear expression.

##### ANSWER

$f(x)=2x^{2}-2x+1$

##### SOLUTION
Let $I=\displaystyle \int{\dfrac{(x-1)dx}{x^2\sqrt{2x^2-2x+1}}}=\displaystyle \int{\dfrac{1/x-1/x^2}{\sqrt{2x^2-2x+1}}}dx$

$=\displaystyle \int{\dfrac{1/x^2-1/x^3}{\sqrt{2-\dfrac{2}{x}+\dfrac{1}{x^2}}}}dx$

put
$2-\dfrac{2}{x}+\dfrac{1}{x^2}=t$

$\left(\dfrac{2}{x^2}-\dfrac{2}{x^3}\right)dx=dt$

$I=\dfrac{1}{2}\displaystyle \int{\dfrac{dt}{\sqrt t}}=\sqrt t+c$
put $t=2-\dfrac{2}{x}+\dfrac{1}{x^2}$

$I=\sqrt{2-\dfrac{2}{x}+\dfrac{1}{x^2}}+c=\dfrac{\sqrt{2x^2-2x+1}}{x}+c$

$f(x)=2x^2-2x+1$

$g(x)=x$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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