The value of int frac{1}{x^2 - 5x + 8} dx is | vedclass

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Question

(A) Step $1$: Complete the square for the denominator $x^2 - 5x + 8 = (x^2 - 5x + \frac{25}{4}) + 8 - \frac{25}{4} = (x - \frac{5}{2})^2 + \frac{7}{4}

Vedclass · Accepted Answer

$\frac{2}{\sqrt{7}} 	an^{-1} \left( \frac{2x - 5}{\sqrt{7}} ight) + c$

Vedclass · Answer

(A) Step $1$: Complete the square for the denominator $x^2 - 5x + 8 = (x^2 - 5x + \frac{25}{4}) + 8 - \frac{25}{4} = (x - \frac{5}{2})^2 + \frac{7}{4}$.Step $2$: The integral becomes $\int \frac{1}{(x - \frac{5}{2})^2 + (\frac{\sqrt{7}}{2})^2} dx$.Step $3$: Use the formula $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + c$.Step $4$: Here $u = x - \frac{5}{2}$ and $a = \frac{\sqrt{7}}{2}$.Step $5$: The result is $\frac{1}{\sqrt{7}/2} 	an^{-1}(\frac{x - 5/2}{\sqrt{7}/2}) + c = \frac{2}{\sqrt{7}} 	an^{-1}(\frac{2x - 5}{\sqrt{7}}) + c$.

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

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