Question

The value of \(\int \frac{1}{x^2-5x+8}dx\) is equal to...

• A. \(\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x+5}{\sqrt{7}})+c\)
• B. \(-\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x-5}{\sqrt{7}})+c\)
• C. \(\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x-5}{\sqrt{7}})+c\)
• D. \(-\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x+5}{\sqrt{7}})+c\)

Hint:

Complete the square in the denominator and use the standard arctangent integral.

Answer 1

The Correct Option is C

Step 1: Differentiate option C to check:
$\dfrac{d}{dx}\left[\dfrac{2}{\sqrt7}\tan^{-1}\dfrac{2x-5}{\sqrt7}\right]=\dfrac{2}{\sqrt7}\cdot\dfrac{2/\sqrt7}{1+\frac{(2x-5)^2}{7}}$.

Step 2: Simplify:
$=\dfrac{4/7}{\frac{7+(2x-5)^2}{7}}=\dfrac{4}{7+4x^2-20x+25}=\dfrac{4}{4x^2-20x+32}=\dfrac{1}{x^2-5x+8}$.

Step 3: Pick:
This equals the integrand, so option C is correct.

Final Answer:
Completing the square gives an arctangent with argument (2x - 5)/sqrt 7. \[ \boxed{\text{(C) }\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{2x-5}{\sqrt7}\right)+c} \]