解答
∫x3−16x2−64x2x−1dx
解答
−821ln821(x−8)+1+821ln821(x−8)−1+641ln∣x∣−12821(2lnx2−16x−64−ln821x−21+1+ln821x−21−1+2ln82x−21+1−2ln82x−21−1)+C
求解步骤
∫x3−16x2−64x2x−1dx
乘开 x3−16x2−64x2x−1:x3−16x2−64x2x−x3−16x2−64x1
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=∫x3−16x2−64x2xdx−∫x3−16x2−64x1dx
∫x3−16x2−64x2xdx=−821(ln82x−8+1−ln82x−8−1)
∫x3−16x2−64x1dx=−641ln∣x∣+641(21lnx2−16x−64−221(ln821x−21+1−ln821x−21−1)+2(21ln821x−21+1−21ln821x−21−1))
=−821(ln82x−8+1−ln82x−8−1)−(−641ln∣x∣+641(21lnx2−16x−64−221(ln821x−21+1−ln821x−21−1)+2(21ln821x−21+1−21ln821x−21−1)))
化简 −821(ln82x−8+1−ln82x−8−1)−(−641ln∣x∣+641(21lnx2−16x−64−221(ln821x−21+1−ln821x−21−1)+2(21ln821x−21+1−21ln821x−21−1))):−821ln821(x−8)+1+821ln821(x−8)−1+641ln∣x∣−12821(2lnx2−16x−64−ln821x−21+1+ln821x−21−1+2ln82x−21+1−2ln82x−21−1)
=−821ln821(x−8)+1+821ln821(x−8)−1+641ln∣x∣−12821(2lnx2−16x−64−ln821x−21+1+ln821x−21−1+2ln82x−21+1−2ln82x−21−1)
解答补常数=−821ln821(x−8)+1+821ln821(x−8)−1+641ln∣x∣−12821(2lnx2−16x−64−ln821x−21+1+ln821x−21−1+2ln82x−21+1−2ln82x−21−1)+C