int frac{6x^2-17x-5}{(x+3)(x-2)^2} dx= | vedclass

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(B) To solve $\int \frac{6x^2-17x-5}{(x+3)(x-2)^2} dx$,we use partial fractions:
$\frac{6x^2-17x-5}{(x+3)(x-2)^2} = \frac{A}{x+3} + \frac{B}{x-2} + \

Vedclass · Accepted Answer

$\log \left| \frac{(x-2)^4}{(x+3)^2} \right| + \frac{3}{x-2} + c$

Vedclass · Answer

(B) To solve $\int \frac{6x^2-17x-5}{(x+3)(x-2)^2} dx$,we use partial fractions:
$\frac{6x^2-17x-5}{(x+3)(x-2)^2} = \frac{A}{x+3} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$.
Multiplying by $(x+3)(x-2)^2$,we get:
$6x^2-17x-5 = A(x-2)^2 + B(x+3)(x-2) + C(x+3)$.
Setting $x=2$: $6(4)-17(2)-5 = C(5) \Rightarrow 24-34-5 = 5C \Rightarrow -15 = 5C \Rightarrow C = -3$.
Setting $x=-3$: $6(9)-17(-3)-5 = A(-5)^2 \Rightarrow 54+51-5 = 25A \Rightarrow 100 = 25A \Rightarrow A = 4$.
Comparing coefficients of $x^2$: $6 = A + B \Rightarrow 6 = 4 + B \Rightarrow B = 2$.
Thus,the integral is $\int \left( \frac{4}{x+3} + \frac{2}{x-2} - \frac{3}{(x-2)^2} \right) dx$.
Integrating term by term: $4 \log |x+3| + 2 \log |x-2| + \frac{3}{x-2} + c$.
This simplifies to $\log |(x+3)^4 (x-2)^2| + \frac{3}{x-2} + c$.
Note: The provided options seem to contain typos. Based on standard partial fraction decomposition,the correct form is $\log |(x+3)^4 (x-2)^2| + \frac{3}{x-2} + c$.

$\int \frac{6x^2-17x-5}{(x+3)(x-2)^2} dx=$

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