Question

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

• A. $f\left(x\right)=2x^2-2x+1$
• B. $g\left(x\right)=x+1$
• C. $g\left(x\right)=x$
• D. $f\left(x\right)=2x^2-2x$

Answer 1

The correct option is A $f\left(x\right)=2x^2-2x+1$

Let $I=\int \frac{(x-1)dx}{x^2\sqrt{2x^2-2x+1}}=\int \frac{1/x-1/x^2}{\sqrt{2x^2-2x+1}}dx$

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

put

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

$(\frac{2}{x^2}-\frac{2}{x^3})dx=dt$

$I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\sqrt{t}+c$

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

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

$f\left(x\right)=2x^2-2x+1$

$g\left(x\right)=x$