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

Exercise

∫x2+13x2−x−4dx\int\frac{x^2+1}{3x^2-x-4}dx

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Final answer to the exercise

13x+43126ln⁡∣6(x−16)7−1∣−43126ln⁡∣6(x−16)7+1∣+118ln⁡∣(x−16)2−4936∣+C0\frac{1}{3}x+\frac{43}{126}\ln\left|\frac{6\left(x-\frac{1}{6}\right)}{7}-1\right|-\frac{43}{126}\ln\left|\frac{6\left(x-\frac{1}{6}\right)}{7}+1\right|+\frac{1}{18}\ln\left|\left(x-\frac{1}{6}\right)^2-\frac{49}{36}\right|+C_0

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dxdy=ey−2y+11+ey\frac{dx}{dy}=e^{\sqrt{y}}-2y+\frac{1}{1+e^y}
∫x2+13x2−x−4 dx
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