Question

Evaluate

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

Answer 1

We have,

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

$\Rightarrow \int \frac{dx}{\sqrt{2(x^2-x+\frac{3}{2})}}$

$\Rightarrow \frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{(x^2-x+\frac{3}{2})}}$

On adding and subtracting denominator ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Then,

$\Rightarrow \frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{(x^2-x+\frac{1}{4}-\frac{1}{4}+\frac{3}{2})}}$

$\Rightarrow \frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{(x^2-x+\frac{1}{4}-\frac{5}{4})}}$

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

Using formula

$\int \frac{dx}{\sqrt{x^2-a^2}}=\log |x+\sqrt{x^2-a^2}|$

$\int \frac{dx}{\sqrt{2x^2-2x+3}}=\log |(x-\frac{1}{2})+\sqrt{{(x-\frac{1}{2})}^2-{\left(\frac{\sqrt{5}}{2}\right)}^2}|$

$=\log |(x-\frac{1}{2})+(x^2-x+\frac{3}{2})|+C$

Hence, this is the answer.