Question:
The value of \(\int \frac{1}{x^2-5x+8}dx\) is equal to...
The value of \(\int \frac{1}{x^2-5x+8}dx\) is equal to...
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Complete the square in the denominator and use the standard arctangent integral.
Updated On: Oct 1, 2026
- \(\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x+5}{\sqrt{7}})+c\)
- \(-\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x-5}{\sqrt{7}})+c\)
- \(\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x-5}{\sqrt{7}})+c\)
- \(-\frac{2}{\sqrt{7}}tan^{-1}(\frac{2x+5}{\sqrt{7}})+c\)
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The Correct Option is C
Solution and Explanation
Step 1: Complete the square:
\(x^2-5x+8=\left(x-\dfrac52\right)^2+8-\dfrac{25}{4}=\left(x-\dfrac52\right)^2+\dfrac74\).
Step 2: Use the standard form:
\(\int\dfrac{dx}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac ua+c\) with \(u=x-\dfrac52\) and \(a=\dfrac{\sqrt7}{2}\).
Step 3: Substitute:
\[ I=\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{x-5/2}{\sqrt7/2}\right)+c=\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{2x-5}{\sqrt7}\right)+c \]
Step 4: Why the other options are wrong.
Options A and D have \(2x+5\), which comes from a sign error in completing the square. Options B and D have a negative sign in front, but the integrand is positive, so the antiderivative must increase.
Final Answer:
The integral is (2/sqrt 7) tan inverse((2x - 5)/sqrt 7) + c. \[ \boxed{\text{(C) }\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{2x-5}{\sqrt7}\right)+c} \]
\(x^2-5x+8=\left(x-\dfrac52\right)^2+8-\dfrac{25}{4}=\left(x-\dfrac52\right)^2+\dfrac74\).
Step 2: Use the standard form:
\(\int\dfrac{dx}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac ua+c\) with \(u=x-\dfrac52\) and \(a=\dfrac{\sqrt7}{2}\).
Step 3: Substitute:
\[ I=\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{x-5/2}{\sqrt7/2}\right)+c=\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{2x-5}{\sqrt7}\right)+c \]
Step 4: Why the other options are wrong.
Options A and D have \(2x+5\), which comes from a sign error in completing the square. Options B and D have a negative sign in front, but the integrand is positive, so the antiderivative must increase.
Final Answer:
The integral is (2/sqrt 7) tan inverse((2x - 5)/sqrt 7) + c. \[ \boxed{\text{(C) }\dfrac{2}{\sqrt7}\tan^{-1}\left(\dfrac{2x-5}{\sqrt7}\right)+c} \]
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