If int {frac{{2x + 3}}{{{x^2} - 5x + 6}}} ;dx | vedclass

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

Question

(C) We are given the integral: $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx$. First,we express the numerator in terms of the derivative of the denom

Vedclass · Accepted Answer

$	ext{Constant}$

Vedclass · Answer

(C) We are given the integral: $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx$. First,we express the numerator in terms of the derivative of the denominator: $\frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$. So,$\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx = \int {\frac{{2x - 5 + 8}}{{{x^2} - 5x + 6}}} \;dx = \int {\frac{{2x - 5}}{{{x^2} - 5x + 6}}} \;dx + \int {\frac{8}{{(x - 2)(x - 3)}}} \;dx$. Using partial fractions for the second part: $\frac{8}{(x - 2)(x - 3)} = \frac{8}{x - 3} - \frac{8}{x - 2}$. Thus,the integral becomes: $\ln |x^2 - 5x + 6| + 8 \int (\frac{1}{x - 3} - \frac{1}{x - 2}) \;dx + C$. $= \ln |(x - 2)(x - 3)| + 8 \ln |x - 3| - 8 \ln |x - 2| + C$. $= \ln |x - 2| + \ln |x - 3| + 8 \ln |x - 3| - 8 \ln |x - 2| + C$. $= 9 \ln |x - 3| - 7 \ln |x - 2| + C$. Comparing this with the given expression $9 \ln (x - 3) - 7 \ln (x - 2) + A$,we find that $A$ must be an arbitrary constant $C$.

If $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx = 9\ln (x - 3) - 7\ln (x - 2) + A$,then $A = $

Similar Questions

$\int \frac{x+1}{x(1+x e^x)^2} \,d x=$

$\int \frac{\tan x}{\cos x(\sec x-1)(\sec x-2)} d x=$ . . . . . . $+c$

If $f(x)$ is a polynomial of the second degree in $x$ such that $f(0)=3, f(1)=3, f(2)=-3$. Then,$\int \frac{f(x)}{x^3-1} d x=$

$\int \frac{dx}{e^x + 1 - 2e^{-x}} = $

$\int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module