解答
∫(4x2+3x+1)3xdx
解答
512−32(64x2+48x+16)21−31367(64x2+48x+16)23(12288arctan(71(8x+3))(x+83)4+15367(x+83)3+2688arctan(71(8x+3))(x+83)2+2807(x+83)+147arctan(71(8x+3)))+C
求解步骤
∫(4x2+3x+1)3xdx
对 4x2+3x+1配方:4(x+83)2+167
=∫(4(x+83)2+167)3xdx
使用换元积分法
=∫(64u2+7)3512(8u−3)du
提出常数: ∫a⋅f(x)dx=a⋅∫f(x)dx=512⋅∫(64u2+7)38u−3du
乘开 (64u2+7)38u−3:(64u2+7)38u−(64u2+7)33
=512⋅∫(64u2+7)38u−(64u2+7)33du
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=512(∫(64u2+7)38udu−∫(64u2+7)33du)
∫(64u2+7)38udu=−32(64u2+7)21
∫(64u2+7)33du=31367(64u2+7)23(12288u4arctan(78u)+15367u3+2688u2arctan(78u)+2807u+147arctan(78u))
=512−32(64u2+7)21−31367(64u2+7)23(12288u4arctan(78u)+15367u3+2688u2arctan(78u)+2807u+147arctan(78u))
u=x+83代回=512−32(64(x+83)2+7)21−31367(64(x+83)2+7)23(12288(x+83)4arctan(78(x+83))+15367(x+83)3+2688(x+83)2arctan(78(x+83))+2807(x+83)+147arctan(78(x+83)))
化简 512−32(64(x+83)2+7)21−31367(64(x+83)2+7)23(12288(x+83)4arctan(78(x+83))+15367(x+83)3+2688(x+83)2arctan(78(x+83))+2807(x+83)+147arctan(78(x+83))):512−32(64x2+48x+16)21−31367(64x2+48x+16)23(12288arctan(71(8x+3))(x+83)4+15367(x+83)3+2688arctan(71(8x+3))(x+83)2+2807(x+83)+147arctan(71(8x+3)))
=512−32(64x2+48x+16)21−31367(64x2+48x+16)23(12288arctan(71(8x+3))(x+83)4+15367(x+83)3+2688arctan(71(8x+3))(x+83)2+2807(x+83)+147arctan(71(8x+3)))
解答补常数=512−32(64x2+48x+16)21−31367(64x2+48x+16)23(12288arctan(71(8x+3))(x+83)4+15367(x+83)3+2688arctan(71(8x+3))(x+83)2+2807(x+83)+147arctan(71(8x+3)))+C