解答
∫(x2+4)2(x+1)32x+1dx
解答
−1000031arctan(2x)−62519lnx2+4+62538ln∣x+1∣−400019sin(2arctan(2x))+250(x2+4)21−125(x+1)6+50(x+1)21+C
求解步骤
∫(x2+4)2(x+1)32x+1dx
将(x2+4)2(x+1)32x+1用部份分式展开: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−19+625(x+1)38+125(x+1)26−25(x+1)31dx
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫625(x2+4)−38x+8dx+∫125(x2+4)2−21x−19dx+∫625(x+1)38dx+∫125(x+1)26dx−∫25(x+1)31dx
∫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))))
∫625(x+1)38dx=62538ln∣x+1∣
∫125(x+1)26dx=−125(x+1)6
∫25(x+1)31dx=−50(x+1)21
=6251(−19lnx2+4+4arctan(2x))+1251(2(x2+4)21−1619(arctan(2x)+21sin(2arctan(2x))))+62538ln∣x+1∣−125(x+1)6−(−50(x+1)21)
化简 6251(−19lnx2+4+4arctan(2x))+1251(2(x2+4)21−1619(arctan(2x)+21sin(2arctan(2x))))+62538ln∣x+1∣−125(x+1)6−(−50(x+1)21):−1000031arctan(2x)−62519lnx2+4+62538ln∣x+1∣−400019sin(2arctan(2x))+250(x2+4)21−125(x+1)6+50(x+1)21
=−1000031arctan(2x)−62519lnx2+4+62538ln∣x+1∣−400019sin(2arctan(2x))+250(x2+4)21−125(x+1)6+50(x+1)21
解答补常数=−1000031arctan(2x)−62519lnx2+4+62538ln∣x+1∣−400019sin(2arctan(2x))+250(x2+4)21−125(x+1)6+50(x+1)21+C