Solving the Integral of 1/(2x² - 2x + 1)

We need to evaluate the integral:

\(\smallint \frac{{dx}}{{2{x^2} - 2x + 1}}\)

This is an integral of the form \(\smallint \frac{{dx}}{{ax^2 + bx + c}}\). To solve this type of integral, we typically complete the square in the denominator.

Let's work on the denominator \(2x^2 - 2x + 1\):

• Factor out the coefficient of \(x^2\), which is 2:

  \(2(x^2 - x + \frac{1}{2})\)

• Complete the square for the quadratic expression inside the parenthesis \(x^2 - x\). To do this, we take half of the coefficient of \(x\) (\(-1\)), square it \((\frac{-1}{2})^2 = \frac{1}{4}\), and add and subtract it inside the parenthesis:

  \(x^2 - x + \frac{1}{4} - \frac{1}{4} + \frac{1}{2}\)

• Group the terms that form a perfect square trinomial:

  \((x^2 - x + \frac{1}{4}) - \frac{1}{4} + \frac{1}{2}\)

• Rewrite the perfect square trinomial as a squared term:

  \((x - \frac{1}{2})^2 + \frac{1}{4}\)

• Now substitute this back into the denominator expression with the factored out 2:

  \(2\left[(x - \frac{1}{2})^2 + \frac{1}{4}\right]\)

• We can rewrite \(\frac{1}{4}\) as \((\frac{1}{2})^2\):

  \(2\left[(x - \frac{1}{2})^2 + (\frac{1}{2})^2\right]\)

So, the denominator is \(2\left[(x - \frac{1}{2})^2 + (\frac{1}{2})^2\right]\). The integral becomes:

\(\smallint \frac{{dx}}{{2\left[(x - \frac{1}{2})^2 + (\frac{1}{2})^2\right]}}\)

We can take the constant \(\frac{1}{2}\) out of the integral:

\(\frac{1}{2} \smallint \frac{{dx}}{{(x - \frac{1}{2})^2 + (\frac{1}{2})^2}}\)

This integral is now in the standard form \(\smallint \frac{{du}}{{u^2 + a^2}}\), where \(u = x - \frac{1}{2}\) and \(a = \frac{1}{2}\). The differential \(du\) is equal to \(dx\) since the derivative of \(x - \frac{1}{2}\) with respect to \(x\) is 1.

The standard integral formula is \(\smallint \frac{{du}}{{u^2 + a^2}} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C\).

Applying this formula with \(u = x - \frac{1}{2}\) and \(a = \frac{1}{2}\):

\(\frac{1}{2} \left[ \frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{x - \frac{1}{2}}{\frac{1}{2}}\right) \right] + C\)

Simplify the expression:

\(\frac{1}{2} \left[ 2 \tan^{-1}\left(\frac{\frac{2x - 1}{2}}{\frac{1}{2}}\right) \right] + C\)

\(\frac{1}{2} \left[ 2 \tan^{-1}(2x - 1) \right] + C\)

\(\tan^{-1}(2x - 1) + C\)

Comparing this result with the given options, we find that it matches option 4:

• Option 1: \(\frac{{{{\tan }^{ - 1}}\left( {2x - 1} \right)}}{2} + c\)
• Option 2: \(2 \tan^{ -1 }(2x - 1) + c\)
• Option 3: \(\frac{{{{\tan }^{ - 1}}\left( {2x + 1} \right)}}{2} + c\)
• Option 4: \(\tan^{ -1 }(2x - 1) + c\)

Therefore, the integral \(\smallint \frac{{dx}}{{2{x^2} - 2x + 1}}\) is equal to \(\tan^{-1}(2x - 1) + c\).