Question

这两题高数题怎么做？

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Answer 1

可以使用换元法

详情如图所示，

有任何疑惑，欢迎追问

Answer 2

解如下图所示

Answer 3

(5)

let
e^x = (tanu)^2
e^x dx = 2(tanu).(secu)^2 du
dx =[2(secu)^2/tanu ] du
∫dx/√(1+e^x)

=∫[2(secu)^2/tanu ] du/secu
=∫[2(secu)/tanu ] du
=2∫ cscu du
=2ln|cscu- cotu| +C
=2ln| √(1+e^x) /e^(x/2)- e^(-x/2)| +C
= 2ln[ √(1+e^x) -1 ] -x + C
(2)

let
x=sinu
dx=cosu du
x=0, u=0
x=1, u=π/2
∫(-1->1) (2x^2+xcosx)/[1+√(1-x^2)] dx

=∫(-1->1) 2x^2/[1+√(1-x^2)] dx +∫(-1->1) xcosx/[1+√(1-x^2)] dx
=∫(-1->1) 2x^2/[1+√(1-x^2)] dx + 0
=4∫(0->1) x^2/[1+√(1-x^2)] dx
=4∫(0->1) [1-√(1-x^2)] dx
=4∫(0->π/2) (1-cosu) ( cosu du )
=2∫(0->π/2) [2cosu -2(cosu)^2 ] du
=2∫(0->π/2) [2cosu -1-cos2u ] du
=2[ 2sinu -u -(1/2)sin2u]|(0->π/2)
=2 ( 2 -π/2)
=4-π

Answer 4

Answer 5

这个是高等数学吗