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

Exercise

∫1−x4x(x2−3x+1)dx\int\frac{1-x}{4x\left(x^2-3x+1\right)}dx

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Final answer to the exercise

14ln⁡∣x∣+5ln⁡∣2(x−32)5−1∣−5ln⁡∣2(x−32)5+1∣40−14ln⁡∣(x−32)2−54∣+C2\frac{1}{4}\ln\left|x\right|+\frac{\sqrt{5}\ln\left|\frac{2\left(x-\frac{3}{2}\right)}{\sqrt{5}}-1\right|-\sqrt{5}\ln\left|\frac{2\left(x-\frac{3}{2}\right)}{\sqrt{5}}+1\right|}{40}-\frac{1}{4}\ln\left|\sqrt{\left(x-\frac{3}{2}\right)^2-\frac{5}{4}}\right|+C_2

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∫1−x4x(x2−3x+1) dx
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