int frac{3x-2}{(x+1)(x-2)^2} dx = (where C i | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{-5}{9} \log |x+1| + \frac{5}{9} \log |x-2| - \frac{4}{3(x-2)} + C$

Vedclass · Answer

(A) To evaluate the integral $\int \frac{3x-2}{(x+1)(x-2)^2} dx$,we use partial fractions. Let $\frac{3x-2}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$. Multiplying by $(x+1)(x-2)^2$,we get $3x-2 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)$. Setting $x = -1$: $3(-1)-2 = A(-1-2)^2 \implies -5 = 9A \implies A = -\frac{5}{9}$. Setting $x = 2$: $3(2)-2 = C(2+1) \implies 4 = 3C \implies C = \frac{4}{3}$. Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = \frac{5}{9}$. Thus,$\int \left( \frac{-5/9}{x+1} + \frac{5/9}{x-2} + \frac{4/3}{(x-2)^2} \right) dx = -\frac{5}{9} \log |x+1| + \frac{5}{9} \log |x-2| - \frac{4}{3(x-2)} + C$.

$\int \frac{3x-2}{(x+1)(x-2)^2} dx = $ (where $C$ is a constant of integration.)

Similar Questions

If $\int \frac{\sin x}{\sin ^{3} x+\cos ^{3} x} d x = \alpha \log _{e}|1+\tan x|+\beta \log _{e}\left|1-\tan x+\tan ^{2} x\right|+\gamma \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C$,where $C$ is the constant of integration,then the value of $18(\alpha+\beta+\gamma^{2})$ is .... .

$\int \frac{d x}{x\left(x^2+1\right)}=$

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

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module