An antiderivative of [math: 2(3x+4)−4] is:

Question

BlackTom AI · Accepted Answer

Use substitution u = 3x + 4 so that du = 3 dx (key idea: u-substitution and the power rule for antiderivatives).

1) Relevant knowledge points and concepts
- Power rule for antiderivatives: ∫u^n du = u^(n+1)/(n+1) for n ≠ −1.
- Chain rule in reverse (u-substitution): when the integrand is f(ax+b), set u = ax+b and account for du/dx = a.

2) Complete reasoning process step by step
Let u = 3x + 4. Then du = 3 dx, so dx = du/3.
The integral of 2(3x+4)^(−4) dx becomes:
∫ 2 u^(−4) (du/3).

3) Calculations with intermediate steps
∫ 2 u^(−4) (du/3) = (2/3) ∫ u^(−4) du.
Apply the power rule: ∫ u^(−4) du = u^(−3)/(−3).
So (2/3) · [u^(−3)/(−3)] + C = (2/3) · (−1/3) u^(−3) + C = −2/9 u^(−3) + C.
Substitute back u = 3x + 4 to get −2/9 (3x + 4)^(−3) + C.

4) Why the correct answer is correct
Differentiating −2/9 (3x+4)^(−3) confirms the result: derivative = −2/9 · (−3)(3x+4)^(−4) · 3 = 2(3x+4)^(−4). The inner derivative factor 3 cancels appropriately with the 1/3 introduced by substitution, producing the exact original integrand.

5) Why the other choices are incorrect
- Choice a: −2/9 (3x+4)^(−5) has the wrong exponent; integrating increases exponent from −4 to −3, not to −5.
- Choice b: −2/3 (3x+4)^(−3) has the wrong coefficient (off by factor 3); it neglects dividing by the inner derivative 3.
- Choice d: −8(3x+4)^(−3) has both the wrong coefficient and does not account correctly for the inner derivative.

6) Conclusion
The antiderivative is −2/9 (3x+4)^(−3) + C, which matches the provided correct answer: −2/9(3x+4)^(−3).

An antiderivative of [math: 2(3x+4)−4] is:单项选择题

类似问题

An antiderivative of \( 2(3x+4)^{-4} \) is:

Find the indefinite integral.

Integrate ∫ ( 2 𝑥 − 1 ) 4 𝑑 𝑥

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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