\[ \int \frac{1+3 x}{(1-x)(3 x-5)} d x= \] (A) \( 2 \ln |1-x|-3 \ln |3 x-5|+C \) (B) \( 2 \ln |1-x|-27 \ln |3 x-5|+C \) (C) \( -2 \ln |1-x|-3 \ln |3 x-5|+C \) (D) \( -2 \ln |1-x|-9 \ln |3 x-5|+C \)
Added by Carol S.
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Step 1: **Partial Fraction Decomposition** We start by expressing the integrand as a sum of partial fractions: \[ \frac{1+3x}{(1-x)(3x-5)} = \frac{A}{1-x} + \frac{B}{3x-5} \] Multiply through by the common denominator \((1-x)(3x-5)\) to clear the Show more…
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Consider the following indefinite integral: I = ∫ (6x^3 + 2x^2 - 46x - 12) / (x^2 - 9) dx The integrand decomposes into the form: ax + b + c/(x - 3) + d/(x + 3) Compute the coefficients a, b, c and d. Evaluate the integral I. A. a = 6, b = 2, c = -5, d = 3, I = 3x^2 + 2x + 5 ln|x - 3| + 3 ln|x + 3| + C B. a = 6, b = 2, c = 5, d = 3, I = 3x^2 + 2x + 5 ln|x - 3| + 3 ln|x + 3| + C C. a = 6, b = 2, c = 5, d = -3, I = 3x^2 + 2x + 5 ln|x - 3| - 3 ln|x + 3| + C D. a = 6, b = -2, c = 5, d = 3, I = 3x^2 + 2x + 5 ln|x - 3| + 3 ln|x + 3| + C
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$\int \frac{\ln x}{3 x} d x=$ (A) $\quad 6 \ln ^{2}|x|+C$ (B) $\quad \frac{1}{3} \ln ^{2}|x|+C$ (C) $\frac{1}{6} \ln ^{2}|x|+C$ (D) $\quad \frac{1}{3} \ln |x|+C$
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