解答
∫(x+1)3⋅(x2+4)22x+1dx
解答
−1000031arctan(2x)+62538ln∣x+1∣−62519lnx2+4−400019sin(2arctan(2x))−125(x+1)6+50(x+1)21+250(x2+4)21+C
求解步骤
∫(x+1)3(x2+4)22x+1dx
将(x+1)3(x2+4)22x+1用部份分式展开:625(x+1)38+125(x+1)26−25(x+1)31+625(x2+4)−38x+8+125(x2+4)2−21x−19
=∫625(x+1)38+125(x+1)26−25(x+1)31+625(x2+4)−38x+8+125(x2+4)2−21x−19dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫625(x+1)38dx+∫125(x+1)26dx−∫25(x+1)31dx+∫625(x2+4)−38x+8dx+∫125(x2+4)2−21x−19dx
∫625(x+1)38dx=62538ln∣x+1∣
∫125(x+1)26dx=−125(x+1)6
∫25(x+1)31dx=−50(x+1)21
∫625(x2+4)−38x+8dx=6251(−19lnx2+4+4arctan(2x))
∫125(x2+4)2−21x−19dx=1251(2(x2+4)21−1619(arctan(2x)+21sin(2arctan(2x))))
=62538ln∣x+1∣−125(x+1)6−(−50(x+1)21)+6251(−19lnx2+4+4arctan(2x))+1251(2(x2+4)21−1619(arctan(2x)+21sin(2arctan(2x))))
化简 62538ln∣x+1∣−125(x+1)6−(−50(x+1)21)+6251(−19lnx2+4+4arctan(2x))+1251(2(x2+4)21−1619(arctan(2x)+21sin(2arctan(2x)))):−1000031arctan(2x)+62538ln∣x+1∣−62519lnx2+4−400019sin(2arctan(2x))−125(x+1)6+50(x+1)21+250(x2+4)21
=−1000031arctan(2x)+62538ln∣x+1∣−62519lnx2+4−400019sin(2arctan(2x))−125(x+1)6+50(x+1)21+250(x2+4)21
解答补常数=−1000031arctan(2x)+62538ln∣x+1∣−62519lnx2+4−400019sin(2arctan(2x))−125(x+1)6+50(x+1)21+250(x2+4)21+C