高等数学，不定积分

Question

不定积分 ∫2^x3^xdx
∫(2/(1+x^2)-4^x-2)dx
∫(3-2x)^2*dx
∫(tanx/cos^2*x)dx
∫xlnxdx
求导 y=3^x-x^2-2^arcsinx y=ln(x+√(x^2+1)) y=cos(1/x) y=√(x+√x)

定积分∫[2 1] x^5*dx
∫[1 0] cos2xdx

极限 lim (x→∞)（x-cosx）/x
lim (x→0) (1-cosx)/x^2
lim (x→∞)(1-3/x)^(-5x-2)

答一题给10分，先给50，后面悬赏

Answer 1

用图片好看，分开三部分做：不定积分

导数：

定积分+极限：

Answer 2

lim (x→∞)（x-cosx）/x =1
lim (x→0) (1-cosx)/x^2=1/2
lim (x→∞)(1-3/x)^(-5x-2)=e^15

Answer 3

∫2^x3^xdx
=∫6^xdx=6^x/ln6+C,
∫(2/(1+x^2)-4^x-2)dx
=2arctanx-4^x/ln4-2x+C,
∫(3-2x)^2*dx =∫(9-12x+4x^2)dx
=9x-6x^2+4x^3/3+C,
∫(tanx/cos^2*x)dx
=∫tanxdtanx
=(tanx)^2/2+C,
∫xlnxdx
=∫lnxd(x^2/2)
=(x^2lnx)/2-∫(x^2/2)dlnx
=(x^2lnx)/2-(1/2)∫xdx
=(x^2lnx)/2-x^2/4+C,
y=3^x-x^2-2^arcsinx
y'=3^xln3-2x-(2^arcsinx)ln2/√(1-x^2),
y=ln(x+√(x^2+1))
y'=[1+(x^2+1)^(-1/2)*2]/[x+√(x^2+1))]
=√(x^2+1),
y=cos(1/x)
y'=-sin(1/x)*(-1/x^2)
=sin(1/x)/x^2,
y=√(x+√x)
y'=(1/2)(x+√x)^(-1/2)[1+1/(2√x)]
=(2√x+1)/(4√(x^2+x√x),
∫[1,2] x^5*dx
=x^6/6[1,2]
=(64-1)/6
=21/2.
∫[0,2] cos2xdx
=(1/2)sin2x[0,2]
=(1/2)sin4,
lim (x→∞)（x-cosx）/x
= 1,
lim (x→0) (1-cosx)/x^2
= lim (x→0)[2sin^2(x/2)/[4*(x/2)^2]
=(1/2) lim (x→0)[sinx/2/(x/2)]^2
=1/2,
lim (x→∞)(1-3/x)^(-5x-2)
设u=3/x,x→∞,u→0,
lim (u→0)(1-u)^(-15/u-2)
v=-u,
= lim (v→0)[(1+v)^1/v]^15/(1+v)^2
=e^15.