해법
적분 (x4−2x2+1)−4x
해법
−xln∣x+1∣+x+1x+xln∣x−1∣+x−1x+(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1)+C
솔루션 단계
∫x4−2x2+1−4xdx
정수를 빼라: ∫a⋅f(x)dx=a⋅∫f(x)dx=−4⋅∫x4−2x2+1xdx
부품별 통합 적용
=−4(x(41ln∣x+1∣−4(x+1)1−41ln∣x−1∣−4(x−1)1)−∫4(x+1)(x−1)x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣dx)
∫4(x+1)(x−1)x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣dx=4(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1)
=−4(x(41ln∣x+1∣−4(x+1)1−41ln∣x−1∣−4(x−1)1)−4(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1))
−4(x(41ln∣x+1∣−4(x+1)1−41ln∣x−1∣−4(x−1)1)−4(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1))간소화하다 :−xln∣x+1∣+x+1x+xln∣x−1∣+x−1x+(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1)
=−xln∣x+1∣+x+1x+xln∣x−1∣+x−1x+(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1)
솔루션에 상수 추가=−xln∣x+1∣+x+1x+xln∣x−1∣+x−1x+(x+1)(x−1)x(x2ln∣x+1∣−x2ln∣x−1∣−2x+ln∣x−1∣−ln∣x+1∣)−(x+11−x−11)(x+1)(x−1)+C