If int frac{x^2}{(x-1)(x-2)(x-3)} dx = log_e | vedclass

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

Question

(B) To solve $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx$,we use partial fractions. Let $\frac{x^2}{(x-1)(x-2)(x-3)} = 1 + \frac{A}{x-1} + \frac{B}{x-2} + \f

Vedclass · Accepted Answer

$C \frac{|x-1|^{1/2} |x-3|^{9/2}}{(x-2)^4}$

Vedclass · Answer

(B) To solve $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx$,we use partial fractions. Let $\frac{x^2}{(x-1)(x-2)(x-3)} = 1 + \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. Using the cover-up rule: $A = \frac{1^2}{(1-2)(1-3)} = \frac{1}{(-1)(-2)} = \frac{1}{2}$. $B = \frac{2^2}{(2-1)(2-3)} = \frac{4}{(1)(-1)} = -4$. $C = \frac{3^2}{(3-1)(3-2)} = \frac{9}{(2)(1)} = \frac{9}{2}$. Thus,the integral becomes $\int (1 + \frac{1/2}{x-1} - \frac{4}{x-2} + \frac{9/2}{x-3}) dx$. $= x + \frac{1}{2} \log_e |x-1| - 4 \log_e |x-2| + \frac{9}{2} \log_e |x-3| + C'$. $= \log_e e^x + \log_e |x-1|^{1/2} - \log_e |x-2|^4 + \log_e |x-3|^{9/2} + C'$. $= \log_e \left( \frac{e^x |x-1|^{1/2} |x-3|^{9/2}}{|x-2|^4} \right) + C'$. Comparing with $\log_e f(x)$,we get $f(x) = C \cdot \frac{|x-1|^{1/2} |x-3|^{9/2}}{(x-2)^4}$.

If $\int \frac{x^2}{(x-1)(x-2)(x-3)} dx = \log_e f(x) + C$,then $f(x) =$

Similar Questions

Integrate the rational function: $\frac{1}{x^{4}-1}$

$\int \frac{6x^2-17x-5}{(x+3)(x-2)^2} dx=$

If $\int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} + 4)}}dx = a\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + b\tan ^{ - 1}\frac{x}{2} + c}$,then the values of $a$ and $b$ are:

$\int \frac{x+1}{x^{2}+5 x+6} d x=$

Evaluate the integral $\int \frac{1}{(x+2)(x^2+1)} \, dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module