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Question at position 3 ∫x2+4x−3x−1dx=\int\frac{x^2+4x-3}{x-1}dx=x22+5x+2ln⁡|x−1|+C\frac{x^2}{2}+5x+2\ln\left|x-1\right|+C7x22−2x+C\frac{7x^2}{2}-2x+Cx22+6x+3ln⁡|x−1|+C\frac{x^2}{2}+6x+3\ln\left|x-1\right|+C12ln⁡|x−1|+C\frac{1}{2}\ln\left|x-1\right|+C13ln⁡|x−1|+C\frac{1}{3}\ln\left|x-1\right|+C

Options
A.𝑥 2 2 + 5 𝑥 + 2 ln ⁡ | 𝑥 − 1 | + 𝐶
B.7 𝑥 2 2 − 2 𝑥 + 𝐶
C.𝑥 2 2 + 6 𝑥 + 3 ln ⁡ | 𝑥 − 1 | + 𝐶
D.1 2 ln ⁡ | 𝑥 − 1 | + 𝐶
E.1 3 ln ⁡ | 𝑥 − 1 | + 𝐶
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Step-by-Step Analysis
We start by examining the given indefinite integral: ∫ (x^2 + 4x − 3) / (x − 1) dx. A practical approach is to perform polynomial long division to simplify the integrand before integrating. Option A (the first choice): x^2/2 + 5x + 2 ln|x−1| + C - This result corresponds to integrating a quotient of (x + 5) with a remainder term 2/(x−1). If we rewrite the integrand after division as (x + 5) + 2/(x−1), then integrating gives ∫(x) dx = x^2/2, ∫(5) dx = 5x, and ∫(2/(x−1)) dx = 2 ln|x−1|. Summing yields x^2/2 + 5x + 2 ln|x−1| + C, which matches this ......Login to view full explanation

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