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f(x)=(3x^2-1)ln(x(1+x)(1-x))−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫(3x2−1)ln⁡(x(1+x)(1−x))dx\int\left(3x^2-1\right)\ln\left(x\left(1+x\right)\left(1-x\right)\right)dx

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Final answer to the exercise

−x3+x3ln⁡∣x∣+ln⁡∣3x+3∣+x3ln⁡∣1+x∣−ln⁡∣−3x+3∣+x3ln⁡∣1−x∣−xln⁡∣x∣+(−x−1)ln⁡∣x+1∣−(−x+1)ln⁡∣−x+1∣−x+C1-x^{3}+x^{3}\ln\left|x\right|+\ln\left|3x+3\right|+x^{3}\ln\left|1+x\right|-\ln\left|-3x+3\right|+x^{3}\ln\left|1-x\right|-x\ln\left|x\right|+\left(-x-1\right)\ln\left|x+1\right|-\left(-x+1\right)\ln\left|-x+1\right|-x+C_1

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(−8.5)0\left(-8.5\right)^0
∫(3x2−1)ln(x(1+x)(1−x))dx
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