MathematicsMedium64×since 2004Q4036 The integral ∫(x8−x2)dx(x12+3x6+1)tan−1(x3+1x3) is equal to : \text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : } The integral ∫(x12+3x6+1)tan−1(x3+x31)(x8−x2)dx is equal to : Aloge(∣tan−1(x3+1x3)∣)1/3+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1 / 3}+\mathrm{C}loge(tan−1(x3+x31))1/3+CBloge(∣tan−1(x3+1x3)∣)+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)+\mathrm{C}loge(tan−1(x3+x31))+CCloge(∣tan−1(x3+1x3)∣)1/2+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1 / 2}+\mathrm{C}loge(tan−1(x3+x31))1/2+CDloge(∣tan−1(x3+1x3)∣)3+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^3+\mathrm{C}loge(tan−1(x3+x31))3+CCheck answerSkip