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Solving: ∫1(w2−4w−3)wdw\int\frac{1}{\left(w^2-4w-3\right)w}dw
Solve instead: ∫1(w2−4w−3)wdx\int\frac{1}{\left(w^2-4w-3\right)w}dx

Exercise

∫1(w2−4w−3)wdx\int\frac{1}{\left(w^2-4w-3\right)w}dx

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Final answer to the exercise

16127ln⁡∣w−2+7∣−16127ln⁡∣w−2−7∣+13ln⁡∣(w−2)2−7∣−13ln⁡∣w∣+C1\frac{16}{127}\ln\left|w-2+\sqrt{7}\right|-\frac{16}{127}\ln\left|w-2-\sqrt{7}\right|+\frac{1}{3}\ln\left|\sqrt{\left(w-2\right)^2-7}\right|-\frac{1}{3}\ln\left|w\right|+C_1

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∫1(w2−4w−3)w dx
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