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

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

BlackTom AI · Accepted Answer

The problem shows a calculus question involving an antiderivative. We need to determine which option matches the correct indefinite integral.
Option A: (2x − 1) / 6 + C. This would correspond to integrating (2x−1)/3 and then incorrectly dividing by 3 again or misapplying constants; its form is not the standard antiderivative of (2x−1)/3. Therefore this is not correct.
Option B: (1/3)(x^2 − x) + C. This matches the straightforward integration: ∫(2x−1)/3 dx = (1/3)∫(2x−1) dx = (1/3)(x^2 − x) + C. This is a correct antiderivative.
Option C: 1/2 + C. This is a constant value, which cannot be the antiderivative of a nonzero function like (2x−1)/3, since differentiating 1/2 would yield 0, not (2x−1)/3. So this is incorrect.
Option D: x^2 − x over something or a garbled expression resembling x^2 − x/3x + C. This form is inconsistent with the proper integral result and appears to mix terms improperly; it does not simplify to (x^2 − x)/3 + C. Thus it is incorrect.
Option E: 2/3 + C. Like Option C, this is a constant and cannot be an antiderivative of a nonzero polynomial integrand; differentiating would give 0, not (2x−1)/3. Therefore it is incorrect.
In summary, among the given choices, only Option B aligns with the correct antiderivative for ∫(2x−1)/3 dx, which evaluates to (x^2 − x)/3 + 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单项选择题

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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 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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