Evaluate the integral: int frac{2x}{4 - 3x - | vedclass

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(C) Let $I = \int \frac{2x}{4 - 3x - x^2} dx$. Factor the denominator: $4 - 3x - x^2 = (4 + x)(1 - x)$. Using partial fractions: $\frac{2x}{(4 + x)(1

Vedclass · Accepted Answer

$-\frac{8}{5} \log |4 + x| - \frac{2}{5} \log |1 - x| + c$

Vedclass · Answer

(C) Let $I = \int \frac{2x}{4 - 3x - x^2} dx$. Factor the denominator: $4 - 3x - x^2 = (4 + x)(1 - x)$. Using partial fractions: $\frac{2x}{(4 + x)(1 - x)} = \frac{A}{4 + x} + \frac{B}{1 - x}$. $2x = A(1 - x) + B(4 + x)$. For $x = 1$: $2 = 5B \implies B = \frac{2}{5}$. For $x = -4$: $-8 = 5A \implies A = -\frac{8}{5}$. $I = \int \left( \frac{-8/5}{4 + x} + \frac{2/5}{1 - x} \right) dx$. $I = -\frac{8}{5} \log |4 + x| + \frac{2/5}{-1} \log |1 - x| + c$. $I = -\frac{8}{5} \log |4 + x| - \frac{2}{5} \log |1 - x| + c$.

Evaluate the integral: $\int \frac{2x}{4 - 3x - x^2} dx$

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