The value of int_2^3 frac{x + 1}{x^2(x - 1)} | vedclass

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

Question

(B) Let $I = \int_2^3 \frac{x + 1}{x^2(x - 1)} dx$. Using partial fractions,we write $\frac{x + 1}{x^2(x - 1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C

Vedclass · Accepted Answer

$\log \frac{16}{9} - \frac{1}{6}$

Vedclass · Answer

(B) Let $I = \int_2^3 \frac{x + 1}{x^2(x - 1)} dx$. Using partial fractions,we write $\frac{x + 1}{x^2(x - 1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{x - 1}$. Multiplying by $x^2(x - 1)$,we get $A(x - 1) + Bx(x - 1) + Cx^2 = x + 1$. Setting $x = 0$,we get $A(-1) = 1 \implies A = -1$. Setting $x = 1$,we get $C(1)^2 = 1 + 1 \implies C = 2$. Comparing the coefficients of $x^2$,we get $B + C = 0 \implies B = -2$. Thus,$I = \int_2^3 \left( -\frac{1}{x^2} - \frac{2}{x} + \frac{2}{x - 1} \right) dx$. $I = \left[ \frac{1}{x} - 2\log|x| + 2\log|x - 1| \right]_2^3$. $I = \left[ \frac{1}{x} + 2\log\left| \frac{x - 1}{x} \right| \right]_2^3$. $I = \left( \frac{1}{3} + 2\log\left( \frac{2}{3} \right) \right) - \left( \frac{1}{2} + 2\log\left( \frac{1}{2} \right) \right)$. $I = \left( \frac{1}{3} - \frac{1}{2} \right) + 2\log\left( \frac{2/3}{1/2} \right) = -\frac{1}{6} + 2\log\left( \frac{4}{3} \right)$. $I = \log\left( \frac{4}{3} \right)^2 - \frac{1}{6} = \log\frac{16}{9} - \frac{1}{6}$.

The value of $\int_2^3 \frac{x + 1}{x^2(x - 1)} dx$ is

Similar Questions

$A$ value of $\alpha$ such that $\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_{e}\left(\frac{9}{8}\right)$ is

The correct evaluation of $\int_0^{\pi /2} {\left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right|} \,dx$ is

$\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx =$

Evaluate the definite integral $\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} dx$.

Let $\{x\}$ denote the fractional part of a real number $x$. Then $\int_0^2 \{x\} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module