int2xlnx^2dx?
回答 (3)
2018-04-07 3:42 pm
Let u = x²
Then, du = 2x dx
∫2x ln(x²) dx
= ∫ln(x²) (2x dx)
= ∫ln(u) du
= u ln(u) - u + C
= x² ln(x²) - x² + C
= 2x² ln(x) - x² + C
Then, du = 2x dx
∫2x ln(x²) dx
= ∫ln(x²) (2x dx)
= ∫ln(u) du
= u ln(u) - u + C
= x² ln(x²) - x² + C
= 2x² ln(x) - x² + C
2018-04-07 7:55 pm
∫ 2x ln x^2 dx
Let t=x^2
dt = 2x dx
∫ 2x ln x^2 dx = ∫ ln t dt
Integrate ∫ ln t dt by parts
dv= dt; v=t
u = ln t ; du = 1/t dt
∫ u dv = uv - ∫ v du
∫ ln t dt = (ln t) t - ∫ t (1/t) dt
∫ ln t dt = (ln t) t - ∫ dt
∫ ln t dt = (ln t) t - t
= t ln t - t
replace t by x^2
= x^2 ln x^2 - x^2 + C
Let t=x^2
dt = 2x dx
∫ 2x ln x^2 dx = ∫ ln t dt
Integrate ∫ ln t dt by parts
dv= dt; v=t
u = ln t ; du = 1/t dt
∫ u dv = uv - ∫ v du
∫ ln t dt = (ln t) t - ∫ t (1/t) dt
∫ ln t dt = (ln t) t - ∫ dt
∫ ln t dt = (ln t) t - t
= t ln t - t
replace t by x^2
= x^2 ln x^2 - x^2 + C
2018-04-07 3:12 pm
No
收錄日期: 2021-05-01 14:46:01
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https://hk.answers.yahoo.com/question/index?qid=20180407071125AAnb9ZC