解答
∫(3x2+12x)52x+4dx
解答
3981312x5−1990656x25+248832x31−248832x41−3981312(x+4)5−1990656(x+4)25−248832(x+4)31−248832(x+4)41+C
求解步骤
∫(3x2+12x)52x+4dx
将(3x2+12x)52x+4用部份分式展开:−3981312x25+995328x35−82944x41+62208x51+3981312(x+4)25+995328(x+4)35+82944(x+4)41+62208(x+4)51
=∫−3981312x25+995328x35−82944x41+62208x51+3981312(x+4)25+995328(x+4)35+82944(x+4)41+62208(x+4)51dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=−∫3981312x25dx+∫995328x35dx−∫82944x41dx+∫62208x51dx+∫3981312(x+4)25dx+∫995328(x+4)35dx+∫82944(x+4)41dx+∫62208(x+4)51dx
∫3981312x25dx=−3981312x5
∫995328x35dx=−1990656x25
∫82944x41dx=−248832x31
∫62208x51dx=−248832x41
∫3981312(x+4)25dx=−3981312(x+4)5
∫995328(x+4)35dx=−1990656(x+4)25
∫82944(x+4)41dx=−248832(x+4)31
∫62208(x+4)51dx=−248832(x+4)41
=−(−3981312x5)−1990656x25−(−248832x31)−248832x41−3981312(x+4)5−1990656(x+4)25−248832(x+4)31−248832(x+4)41
化简=3981312x5−1990656x25+248832x31−248832x41−3981312(x+4)5−1990656(x+4)25−248832(x+4)31−248832(x+4)41
解答补常数=3981312x5−1990656x25+248832x31−248832x41−3981312(x+4)5−1990656(x+4)25−248832(x+4)31−248832(x+4)41+C