Question

The value of \(∫\frac{dx}{x^2-6x+13}\)is:

• A. \(\frac{1}{2} tan^{-1}\frac{x-3}{2}+C\), where C is constant of integration.
• B. \(\frac{1}{2} cot^{-1}\frac{x-3}{2}+C\), where C is constant of integration.
• C. \(\frac{1}{2} tan^{-1}\frac{x+3}{2}+C\), where C is constant of integration.
• D. \(\frac{1}{2} cot^{-1}\frac{x+3}{2}+C\), where C is constant of integration.

Answer 1

The Correct Option is A

To solve the integral \(∫\frac{dx}{x^2-6x+13}\), we need to simplify the expression in the denominator to identify a suitable substitution or method for integration. The quadratic expression \(x^2-6x+13\) can be rewritten by completing the square:

\(x^2-6x+13=(x^2-6x+9)+4=(x-3)^2+4\).

Thus, the integral becomes:

\(∫\frac{dx}{(x-3)^2+4}\).

This resembles the standard integral formula for the inverse tangent function:

\(∫\frac{dx}{a^2+x^2}=\frac{1}{a}tan^{-1}\left(\frac{x}{a}\right)+C\).

In our case, \(a^2=4\) implies \(a=2\), and we have \(x\) replaced by \(x-3\). Applying the formula results in:

\(\frac{1}{2}tan^{-1}\left(\frac{x-3}{2}\right)+C\), where \(C\) is the constant of integration.

Therefore, the value of the integral is:

\(\frac{1}{2}tan^{-1}\frac{x-3}{2}+C\), where \(C\) is the constant of integration.