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(2)
∫(x^3-3x^2+2x)/x^2 dx
=∫(x-3+2/x) dx
=(1/2)x^2-3x +2ln|x| +C
(3)
∫xe^x dx
=∫x de^x
=xe^x -∫e^x dx
=xe^x -e^x +C
(4)
let
u=√x
2udu =dx
x=0,u=0
x=1, u=1
∫(0->1) e^(√x) dx
=∫(0->1) 2u e^u du
=2∫(0->1) u de^u
=2[ue^u]|(0->1) 2∫(0->1) e^u du
=2e-2[e^u]|(0->1)
=2
(5)
let
x^(1/4) =√2tanu
(1/4)x^(-3/4) dx = √2(secu)^2 du
dx =4√2 (tanu)^3.(secu)^2 du
∫ dx/(2+√x)
=∫ 4√2 (tanu)^3.(secu)^2 du/[2(secu)^2]
=2√2 ∫ (tanu)^3 du
=2√2 ∫ [(secu)^2 -1).tanu du
=2√2 [ ∫ secu dsecu -∫ tanu du ]
=2√2 [ (1/2)(secu)^2 +ln|cosu| ] +C
=√2(secu)^2 +2√2.ln|cosu| +C
=√2 [ √x/2 +1] +2√2.ln| 1/√(1+√x) | +C
=(√2/2)x +√2 -√2.ln|1+√x| +C
=(√2/2)x -√2.ln|1+√x| +C'
where
x^(1/4) =√2tanu
∫(x^3-3x^2+2x)/x^2 dx
=∫(x-3+2/x) dx
=(1/2)x^2-3x +2ln|x| +C
(3)
∫xe^x dx
=∫x de^x
=xe^x -∫e^x dx
=xe^x -e^x +C
(4)
let
u=√x
2udu =dx
x=0,u=0
x=1, u=1
∫(0->1) e^(√x) dx
=∫(0->1) 2u e^u du
=2∫(0->1) u de^u
=2[ue^u]|(0->1) 2∫(0->1) e^u du
=2e-2[e^u]|(0->1)
=2
(5)
let
x^(1/4) =√2tanu
(1/4)x^(-3/4) dx = √2(secu)^2 du
dx =4√2 (tanu)^3.(secu)^2 du
∫ dx/(2+√x)
=∫ 4√2 (tanu)^3.(secu)^2 du/[2(secu)^2]
=2√2 ∫ (tanu)^3 du
=2√2 ∫ [(secu)^2 -1).tanu du
=2√2 [ ∫ secu dsecu -∫ tanu du ]
=2√2 [ (1/2)(secu)^2 +ln|cosu| ] +C
=√2(secu)^2 +2√2.ln|cosu| +C
=√2 [ √x/2 +1] +2√2.ln| 1/√(1+√x) | +C
=(√2/2)x +√2 -√2.ln|1+√x| +C
=(√2/2)x -√2.ln|1+√x| +C'
where
x^(1/4) =√2tanu
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