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

Exercise

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

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Final answer to the exercise

−73arctan⁡(2x3)78+213ln⁡∣4x2+3∣+526arctan⁡(x2)+213ln⁡∣2x2+4∣+C1\frac{-7\sqrt{3}\arctan\left(\frac{2x}{\sqrt{3}}\right)}{78}+\frac{2}{13}\ln\left|\sqrt{4x^2+3}\right|+\frac{5}{26}\arctan\left(\frac{x}{2}\right)+\frac{2}{13}\ln\left|\frac{2}{\sqrt{x^2+4}}\right|+C_1

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∫x2+2x−1(4x2+3)(x2+4) dx
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