int1/(root(3)(1-6x))dx=?

1/6(1-6x)^(3/2)+c

1/2(1-6x)^(3/2)+c

1/4(1-6x)^(3/2)+c

-1/4(1-6x)^(2/3)+c

1/4(1-6x)^(2/3)+c

সঠিক উত্তরঃ D.

-1/4(1-6x)^(2/3)+c

Explanation:

Another Explanation (5): ```html

প্রশ্ন: ∫ 1/∛(1-6x) dx = ?

সমাধান:

ধরি, u = 1 - 6x

∴ du = -6 dx

বা, dx = -du/6

এখন, ∫ 1/∛(1-6x) dx = ∫ 1/∛u (-du/6)

= -1/6 ∫ u^-1/3 du

= -1/6 * (u^{(-1/3 + 1)} / (-1/3 + 1)) + c

= -1/6 * (u^(2/3) / (2/3)) + c

= -1/6 * (3/2) * u^2/3 + c

= -1/4 * u^2/3 + c

u এর মান বসিয়ে পাই,

= -1/4 * (1-6x)^2/3 + c

অতএব, ∫ 1/∛(1-6x) dx = -1/4(1-6x)^2/3 + c 😃

সুতরাং, উত্তর: -1/4(1-6x)^2/3 + c 👍

```