Question

Solve $\int \frac{dx}{\sqrt{5x^2-2x}}$

• A. $I=\frac{1}{\sqrt{5}}\log \left|x-\frac{1}{5}-\frac{1}{\sqrt{5}}\sqrt{5x^2-2x}\right|+c$
• B. $I=-\frac{1}{\sqrt{5}}\log \left|x-\frac{1}{5}+\frac{1}{\sqrt{5}}\sqrt{5x^2-2x}\right|+c$
  
• C. $I=\frac{1}{\sqrt{5}}\log \left|x-\frac{1}{5}+\frac{1}{\sqrt{5}}\sqrt{5x^2-2x}\right|+c$
  
• D. None of these

Answer 1

Answer: (C)

$I=\frac{1}{\sqrt{5}}\log \left|x-\frac{1}{5}+\frac{1}{\sqrt{5}}\sqrt{5x^2-2x}\right|+c$

${\int }^{}\frac{dx}{\sqrt{5x^2-2x}}$

$={\int }^{}\frac{1}{\sqrt{5}}\frac{dx}{\sqrt{x^2-\frac{2}{5}x}}$

$=\frac{1}{\sqrt{5}}{\int }^{}\frac{dx}{\sqrt{x^2-\frac{2}{5}x+\frac{1}{25}-\frac{1}{25}}}$

$=\frac{1}{\sqrt{5}}{\int }^{}\frac{dx}{\sqrt{{\left(x-\frac{1}{5}\right)}^2-{\left(\frac{1}{5}\right)}^2}}$

$=\frac{1}{\sqrt{5}}\log \left|x-\frac{1}{5}+\sqrt{x^2-\frac{2}{5}x}\right|+c$

$=\frac{1}{\sqrt{5}}\log \left|x-\frac{1}{5}+\frac{1}{\sqrt{5}}\sqrt{5x^2-2x}\right|+c$