Integrate the rational function: frac{3x-1}{( | vedclass

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

Question

(A) Let $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.$3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.For $x=1$: $3(

Vedclass · Accepted Answer

$\log |x-1| - 5\log |x-2| + 4\log |x-3| + C$

Vedclass · Answer

(A) Let $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.$3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.For $x=1$: $3(1)-1 = A(1-2)(1-3) \Rightarrow 2 = A(-1)(-2) \Rightarrow 2A = 2 \Rightarrow A=1$.For $x=2$: $3(2)-1 = B(2-1)(2-3) \Rightarrow 5 = B(1)(-1) \Rightarrow -B = 5 \Rightarrow B=-5$.For $x=3$: $3(3)-1 = C(3-1)(3-2) \Rightarrow 8 = C(2)(1) \Rightarrow 2C = 8 \Rightarrow C=4$.Thus,$\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3}$.Integrating both sides with respect to $x$:$\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx$.$= \log |x-1| - 5 \log |x-2| + 4 \log |x-3| + C$.

Integrate the rational function: $\frac{3x-1}{(x-1)(x-2)(x-3)}$

Similar Questions

If $\int {\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}dx = {{\log }_e}\left\{ {{{(x - 1)}^{\frac{5}{2}}}{{({x^2} + 1)}^a}} \right\}} - \frac{1}{2}{\tan ^{ - 1}}x + A$,where $A$ is any arbitrary constant,then the value of $a$ is

If $\int \frac{2 x^2}{\left(2 x^2+\alpha\right)\left(x^2+5\right)} d x=\frac{\sqrt{5}}{3} \tan ^{-1} \frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} \tan ^{-1} \frac{x}{\sqrt{2}}+c$,then $\alpha=$

If $729 \int_1^3 \frac{1}{x^3(x^2+9)^2} dx = a + \log b$,then $a - b =$

$\int \frac{3x+4}{x^3-2x-4} dx = \log f(x) + C \Rightarrow f(3) = ?$

$\int \frac{\sin 2x}{\sin^2 x + 3\cos x - 3} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module