TÍNH NGUYÊN HÀM $ \INT {{X^2}\SQRT {{X^3} + 1} \;DX}$ BẰNG CÁCH ĐẶT $ T = \SQRT {{X^3} + 1} .$ MỆNH ĐỀ NÀO SAU ĐÂY LÀ ĐÚNG?
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Level:
Medium
Question Types: Multiple choice
Answers:
- $ \\\int {{x^2}\\\sqrt {{x^3} + 1} \\\;dx} = \\\int {\\\frac{{2{t^2}}}{3}\\\;dt.}$
- $ \\\int {{x^2}\\\sqrt {{x^3} + 1} \\\;dx} = \\\int {\\\frac{{2{t^2}}}{3}\\\;dx.}$
- $ \\\int {{x^2}\\\sqrt {{x^3} + 1} \\\;dx} = \\\int {\\\frac{t}{3}\\\;dt.}$
- $ \\\int {{x^2}\\\sqrt {{x^3} + 1} \\\;dx} = \\\int {{t^3}\\\;dt.}$