Đáp án: tìm nguyên hàm a ∫2x(1+3x^3)dx b fx=(x-1)/(2x-1) - (l)a) ∫2x(1 + 3x^3)dx = 2∫ xdx + 6∫ x^4dx = 2⋅(x^2/2) + 6⋅(x^5/5) + C = x^2 + (6x^5/5) + C b) ∫(x-1/2x-1)dx = ∫[12 - (1/2(2x-1))]dx = 12∫ dx - 12∫(1/2x-1)dx = 12x - 14∫(d(2x-1)/2x-1) = 12x-14ln∣2x-1∣ + C

(l)a) ∫2x(1 + 3x^3)dx = 2∫ xdx + 6∫ x^4dx = 2⋅(x^2/2) + 6⋅(x^5/5) + C = x^2 + (6x^5/5) + C b) ∫(x-1/2x-1)dx = ∫[12 - (1/2(2x-1))]dx = 12∫ dx - 12∫(1/2x-1)dx = 12x - 14∫(d(2x-1)/2x-1) = 12x-14ln∣2x-1∣ + C

trantrunghien1

2 năm trước

tìm nguyên hàm: a. ∫2x(1+3x^3)dx b. f(x)=(x-1)/(2x-1)

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2 năm trước

$\begin{array}{l}a)\quad \displaystyle\int2x(1 + 3x^3)dx\\ = 2\displaystyle\int xdx + 6\displaystyle\int x^4dx\\ = 2\cdot\dfrac{x^2}{2} + 6\cdot\dfrac{x^5}{5} + C\\ = x^2 + \dfrac{6x^5}{5} + C\\ b)\quad \displaystyle\int\dfrac{x-1}{2x-1}dx\\ = \displaystyle\int\left[\dfrac12 - \dfrac{1}{2(2x-1)}\right]dx\\ = \dfrac12\displaystyle\int dx - \dfrac12\displaystyle\int\dfrac{1}{2x-1}dx\\ = \dfrac12x - \dfrac14\displaystyle\int\dfrac{d(2x-1)}{2x-1}\\ = \dfrac12x-\dfrac14\ln|2x-1| + C \end{array}$