int frac{x}{x^3-3 x+2} d x= | vedclass

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

Question

(D) First,factor the denominator: $x^3-3x+2 = (x-1)^2(x+2)$.Using partial fractions: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \fr

Vedclass · Accepted Answer

$-\frac{1}{3} \frac{1}{(x-1)}+\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|+c$

Vedclass · Answer

(D) First,factor the denominator: $x^3-3x+2 = (x-1)^2(x+2)$.Using partial fractions: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.Multiplying by $(x-1)^2(x+2)$ gives: $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$.Setting $x=1$: $1 = B(3) \Rightarrow B = \frac{1}{3}$.Setting $x=-2$: $-2 = C(-3)^2 \Rightarrow -2 = 9C \Rightarrow C = -\frac{2}{9}$.Comparing coefficients of $x^2$: $0 = A + C \Rightarrow A = -C = \frac{2}{9}$.Thus,the integral is: $\int \left( \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2} \right) dx$.$= \frac{2}{9} \log |x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \log |x+2| + c$.$= -\frac{1}{3(x-1)} + \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| + c$.

$\int \frac{x}{x^3-3 x+2} d x=$

Similar Questions

$\int \frac{dx}{x(x^2+1)^3} = ?$

Find $\int \frac{x^{2}+x+1}{(x+2)(x^{2}+1)} dx$.

Integrate the rational function: $\frac{5x}{(x+1)(x^2-4)}$

If $\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx = A \log |x-1| + B \log |x-2| + C \log |x-3| + K$,then the values of $A, B$,and $C$ are respectively:

Evaluate the integral: $\int \frac{x^2 \, dx}{(x^2 + 2)(x^2 + 5)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module