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Exercise

∫0x(t2ln(1+2t))dt\int_0^x\left(t^2ln\left(1+2t\right)\right)dt

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Final answer to the exercise

x3ln⁡∣1+2x∣3−03ln⁡∣1+2⋅0∣3−0.256ln⁡∣3∣+0.256ln⁡∣6x+3∣−0.56x−(−112)x2+−x39\frac{x^{3}\ln\left|1+2x\right|}{3}- \frac{0^{3}\ln\left|1+2\cdot 0\right|}{3}-\frac{0.25}{6}\ln\left|3\right|+\frac{0.25}{6}\ln\left|6x+3\right|-\frac{0.5}{6}x- \left(-\frac{1}{12}\right)x^2+\frac{-x^{3}}{9}

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∫0x(t2ln(1+2t))dt
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