∫(x)/(x+9) dx, u = x+9
Please help!
Step by step if possible.
Indefinite integrals?
2016-06-20 7:51 pm
回答 (2)
2016-06-20 7:57 pm
Let u = x + 9
Then, x = u - 9 and dx = du
∫[x/(x + 9)] dx
= ∫[(u - 9)/u] du
= ∫[1 - (9/u)] du
= ∫du - ∫(9/u) du
= u - 9 ln|u| + C
= x + 9 - 9 ln|x + 9| + C
Then, x = u - 9 and dx = du
∫[x/(x + 9)] dx
= ∫[(u - 9)/u] du
= ∫[1 - (9/u)] du
= ∫du - ∫(9/u) du
= u - 9 ln|u| + C
= x + 9 - 9 ln|x + 9| + C
2016-06-20 8:01 pm
∫(x)/(x+9) dx
u=x+9
du = dx
x = u-9
∫(x)/(x+9) dx = ∫(u-9) / u du
= ∫ du - 9 ∫ du / u
= u - 9 ln |u|
= (x+9) - 9 ln |x+9| + C
u=x+9
du = dx
x = u-9
∫(x)/(x+9) dx = ∫(u-9) / u du
= ∫ du - 9 ∫ du / u
= u - 9 ln |u|
= (x+9) - 9 ln |x+9| + C
收錄日期: 2021-04-18 15:10:28
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