I am given the following
∫(sinx)/(√2+cosx) dx
∫(e^2x)/(e^2x+1)^2 dx
I was able to find the U's for both of the integrals, and I found du for the first one, but then after that I'm stumped.
U substitution for fractions?
2016-05-01 9:01 pm
回答 (2)
2016-05-01 9:21 pm
let u=√2+cos(x)
du=-sin(x) dx
∫(sinx)/(√2+cosx) dx
= - ∫-sin(x) dx / u
= - ∫ du/u
= - ln|u| + C
= - ln|√2+cos(x)| + C
∫ du/u = ln|u| + C
let u=e^(2x) + 1
du = 2e^(2x) dx = 2e^(2x) dx
∫(e^2x)/(e^2x+1)^2 dx
= ½ ∫ 2e^(2x) dx / u²
= ½ ∫ du/u² = ½ ∫ u⁻² du
= ½ u⁻¹ /(-1) + C
= -½ 1/u + C
= -1/[2(e^(2x)+1)] + C
∫ u^(-2) du = u^(-1)/-1 + C
= -1/u + C ; power rule
du=-sin(x) dx
∫(sinx)/(√2+cosx) dx
= - ∫-sin(x) dx / u
= - ∫ du/u
= - ln|u| + C
= - ln|√2+cos(x)| + C
∫ du/u = ln|u| + C
let u=e^(2x) + 1
du = 2e^(2x) dx = 2e^(2x) dx
∫(e^2x)/(e^2x+1)^2 dx
= ½ ∫ 2e^(2x) dx / u²
= ½ ∫ du/u² = ½ ∫ u⁻² du
= ½ u⁻¹ /(-1) + C
= -½ 1/u + C
= -1/[2(e^(2x)+1)] + C
∫ u^(-2) du = u^(-1)/-1 + C
= -1/u + C ; power rule
2016-05-01 9:15 pm
let u = cosx, du = -sinxdx ==> int{-1/(2^0.5 +u) =-ln(2^0.5 + cosx) for the first
for the second, let u = e^2x + 1, du = e^2x dx
int{1/u^2}
-u^3/3, ==>- (e^2x + 1)^3/3
for the second, let u = e^2x + 1, du = e^2x dx
int{1/u^2}
-u^3/3, ==>- (e^2x + 1)^3/3
收錄日期: 2021-04-21 18:27:49
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