x3+x2/x2+4的不定积分

更新时间：2020-01-23 13:10:49 585次访问

∫ ( x^3 + x^2 )/( x^2 + 4 ) dx

= ∫[ ( x^3 + 4x ) + (x^2 + 4 ) - 4x - 4 ] /( x^2 + 4 ) dx

= ∫ ( x + 1 ) dx - 2∫2x/( x^2 + 4 ) dx - ∫ 4/( x^2 + 4 ) dx

= x^2/2 + x - 2∫ d( x^2 + 4 )/( x^2 + 4 ) - 2∫d(x/2)/[ (x/2)^2 + 1 ]

= x^2/2 + x - 2ln( x^2 + 4 ) - 2arctan(x/2) + C 。

∫[x³+x²/(x²+4)]dx

=∫x³dx+∫dx-4∫dx/(x²+4)

=x⁴/4+x-2∫d(x/2)/[(x/2)²+1]

=x⁴/4+x-2arctan(x/2)+c

【如果】

∫[(x³+x²)/(x²+4)]dx

那么

原式

=∫x³dx/(x²+4)+∫x²dx/(x²+4)

=∫xdx-4∫xdx/(x²+4)+∫dx-4∫dx/(x²+4)

=x²/2+x-2ln(x²+4)-2arctan(x/2)+c

∫x²/(x²+4)²dx=-1/2·∫xd[1/(x²+4)]=-x/[2(x²+4)]+1/2·∫1/(x²+4)dx=-x/[2(x²+4)]+1/4·arctan(x/2)+C