The value of {1}{x²-5 x+8} d x is equal to...

Correct Answer is : 27tan⁡−1(2x−57)+c{2}{{7}} ^{-1}≤ft({2 x-5}{{7}})+c72tan−1(72x−5)+c

{2}{{7}} ^{-1}≤ft({2 x+5}{{7}})+c

-{2}{{7}} ^{-1}≤ft({2 x-5}{{7}})+c

{2}{{7}} ^{-1}≤ft({2 x-5}{{7}})+c

-{2}{{7}} ^{-1}≤ft({2 x+5}{{7}})+c

Correct Answer is : 27tan⁡−1(2x−57)+c{2}{{7}} ^{-1}≤ft({2 x-5}{{7}})+c72tan−1(72x−5)+c

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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 129 of 150

Marks: +1, -0

The value of

\int \frac{1}{x^{2}-5 x+8} d x

is equal to...

Solution:

Concept:

The integral is of the form

\int \frac{1}{x^2 - ax + b}\,dx

, which simplifies by completing the square and using the standard

\tan^{-1}

formula.

Explanation:

Complete the square in the denominator:

x^2 - 5x + 8 = \left(x - \frac{5}{2}\right)^2 + \frac{7}{4}

So the integral becomes:

\int \frac{dx}{\left(x - \frac{5}{2}\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2}

Using the standard formula:

\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c

Here,

a = \frac{\sqrt{7}}{2}

, so:

\frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{x - \frac{5}{2}}{\frac{\sqrt{7}}{2}}\right) + c

Simplify the argument:

\frac{x - \frac{5}{2}}{\frac{\sqrt{7}}{2}} = \frac{2x - 5}{\sqrt{7}}

Hence:

\int \frac{1}{x^2 - 5x + 8}\,dx = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{2x - 5}{\sqrt{7}}\right) + c

Answer:

Option C

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