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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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...

Similar Questions

Integrate the function: $\frac{1}{\sqrt{9-25 x^{2}}}$

The velocity of a particle at time $t$ is given by the relation $v = 6t - \frac{t^2}{6}$. Its displacement $S$ is zero at $t = 0$,then the distance travelled in $3 \text{ s}$ is

Find the anti-derivative $F$ of $f$ defined by $f(x) = 4x^{3} - 6$,where $F(0) = 3$.

If the primitive of $f(x) = \pi \sin(\pi x) + 2x - 4$ has the value $3$ for $x = 1$, then the set of $x$ for which the primitive of $f(x)$ vanishes is:

If $\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}$ such that $f(2)=0$,then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module