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

$\int \frac{x - 1}{(x - 3)(x - 2)} \, dx = $

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

Let $I(x) = \int \frac{(x+1)}{x(1+x e^x)^2} dx, x > 0$. If $\lim_{x \rightarrow \infty} I(x) = 0$,then $I(1)$ is equal to

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module