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

Question

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

BlackTom AI · Accepted Answer

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 option exactly. Therefore Option A is consistent with the correct decomposition and integration.

Option B (the second choice): 7x^2/2 − 2x + C
- This corresponds to integrating a quadratic term with a linear term and constant, but there is no justification for obtaining a 7x^2/2 or a −2x from the division step. The original integrand divided by (x−1) does not produce a leading term 7x^2/2; thus this option misrepresents the polynomial part and ignores the logarithmic contribution from the remainder. It’s inconsistent with the actual partial-fraction/quotient form.

Option C (the third choice): x^2/2 + 6x + 3 ln|x−1| + C
- Here, the linear term is 6x instead of 5x, and the coefficient of the logarithm is 3 instead of 2. If we attempt to differentiate this candidate to recover the integrand, we would obtain x + 6 + 3/(x−1), which does not simplify to (x^2 + 4x − 3)/(x − 1). This reveals a mismatch in both the polynomial part and the logarithmic part, indicating incorrect coefficients for both the linear and logarithmic terms.

Option D (the fourth choice): 1/2 ln|x−1| + C
- This option only includes a logarithmic term with coefficient 1/2 and omits the polynomial terms entirely. Differentiating 1/2 ln|x−1| gives 1/(2(x−1)), which does not reproduce the original rational function (x^2 + 4x − 3)/(x−1). Therefore this option is far from correct as it neglects the polynomial contributions that arise from the quotient part of the division.

Option E (the fifth choice): 1/3 ln|x−1| + C
- Similar to Option D, this presents only a logarithmic term, with coefficient 1/3. Differentiating yields 1/(3(x−1)), which again fails to recreate the original integrand. This option also ignores the polynomial components that must come from the division, making it incorrect.

Summary of reasoning:
- The correct method is to perform long division: (x^2 + 4x − 3) ÷ (x − 1) = x + 5 + 2/(x − 1).
- Integrating term-by-term gives ∫x dx = x^2/2, ∫5 dx = 5x, and ∫2/(x − 1) dx = 2 ln|x−1|, plus C.
- Among the options, only the first option matches this result exactly: x^2/2 + 5x + 2 ln|x−1| + C. The other options fail because they misstate the polynomial coefficients or omit the logarithmic part, or both.

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Question at position 11 find ∫23−xdx\int2^{3-x}dx−e3−xln⁡(2)+C- \frac{e^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C \frac{2^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C\frac{2^{3 - x}}{\ln(2)} + C−23−xln⁡(2)+C- \frac{2^{3 - x}}{\ln(2)} + C

Question at position 7 ∫(x3−1x4+2)dx=\int\left(x^3-\frac{1}{x^4}+2\right)dx=x44−3x3+2x+C\frac{x^4}{4}-\frac{3}{x^3}+2x+Cx44+13x3+2x+C\frac{x^4}{4}+\frac{1}{3x^3}+2x+C3x2+4x−5+C3x^2+4x^{-5}+C3x2−14x3+C3x^2-\frac{1}{4x^3}+Cx44−13x3+2x+C\frac{x^4}{4}-\frac{1}{3x^3}+2x+C

Question at position 4 ∫2x−13dx=\int\frac{2x-1}{3}dx=(2x−1)26+C\frac{\left(2x-1\right)^2}{6}+C13(x2−x)+C\frac{1}{3}\left(x^2-x\right)+C12+C\frac{1}{2}+Cx2−x3x+C\frac{x^2-x}{3x}+C23+C\frac{2}{3}+C

Question at position 7 ∫(x3−1x4+2)dx=\int\left(x^3-\frac{1}{x^4}+2\right)dx=x44−13x3+2x+C\frac{x^4}{4}-\frac{1}{3x^3}+2x+Cx44+13x3+2x+C\frac{x^4}{4}+\frac{1}{3x^3}+2x+Cx44−3x3+2x+C\frac{x^4}{4}-\frac{3}{x^3}+2x+C3x2+4x−5+C3x^2+4x^{-5}+C3x2−14x3+C3x^2-\frac{1}{4x^3}+C

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