Integral Calculus

Q1 : ∫ x sin x2 dx



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Q2 : ∫(x + 1)(ln x + 3) dx


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Q1.
∫ x sin(x^2)dx
=∫ sin(x^2)(1/2dx^2)
=1/2∫ sin(x^2)d(x^2)
=-1/2(cos(x^2))+C
Q2.
∫(x+1)(lnx+3)dx
=∫[3(x+1)+x．lnx+lnx]dx
=∫3(x+1)dx + ∫x．lnxdx+∫lnxdx
=(3x^2/2 + 3x + C1) +∫x．lnxdx + ∫lnxdx
=(3x^2/2+3x+C1)+1/2∫lnxdx^2 +∫lnxdx
=(3x^2/2 + 3x + C1) +1/2[x^2．lnx-∫x^2dlnx] +[ x．lnx-∫xdlnx]
=(3x^2/2 + 3x + C1) +1/2(x^2．lnx -∫xdx)+[x．lnx -∫dx](because dlnx/dx=1/x)
=(3/2)x^2 + 3x + (1/2)x^2．lnx - (1/4)x^2+ x．lnx -x + C

Q1.

∫ x sin(x^2)dx

=∫ sin(x^2)(1/2dx^2)

=1/2∫ sin(x^2)d(x^2)

=-1/2(cos(x^2))+C
Q2.
∫(x+1)(lnx+3)dx
=∫[3(x+1)+x．lnx+lnx]dx
=∫3(x+1)dx + ∫x．lnxdx+∫lnxdx
=(3x^2/2 + 3x + C1) +∫x．lnxdx + ∫lnxdx
=(3x^2/2+3x+C1)+1/2∫lnxdx^2 +∫lnxdx
using by part
=(3x^2/2 + 3x + C1) +1/2[x^2．lnx-∫x^2dlnx] +[ x．lnx-∫xdlnx]
=(3x^2/2 + 3x + C1) +1/2(x^2．lnx -∫xdx)+[x．lnx -∫dx](because dlnx/dx=1/x)
=(3/2)x^2 + 3x + (1/2)x^2．lnx - (1/4)x^2+ x．lnx -x + C


2008-01-16 18:04:51 補充：
排好些是:(5/4)ｘ＾２＋ 2x＋(ｘ＾２／２＋ｘ ）．lnｘ＋ｃ

2008-01-16 18:06:31 補充：
（５／４）ｘ＾２＋ ２ｘ＋（ｘ＾２／２＋ｘ ）．ｌｎｘ＋ｃ

Q2) by parts
dv = (x+1)dx
u = (lnx+3)
v = x^2/2+x

∫udv = uv- ∫vdu
= ............
你自己應該識計