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

Exercise

∫(4⋅x+3(x2+2⋅x+5)⋅(x2+4⋅x+5))dx\int\left(\frac{4\cdot x+3}{\left(x^2+2\cdot x+5\right)\cdot\left(x^2+4\cdot x+5\right)}\right)dx

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Final answer to the exercise

1720arctan⁡(x+12)−310ln⁡∣(x+1)2+4∣−75arctan⁡(x+2)+320ln⁡∣(x+2)2+1∣+C1\frac{17}{20}\arctan\left(\frac{x+1}{2}\right)-\frac{3}{10}\ln\left|\sqrt{\left(x+1\right)^2+4}\right|-\frac{7}{5}\arctan\left(x+2\right)+\frac{3}{20}\ln\left|\left(x+2\right)^2+1\right|+C_1

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∫(4·x+3(x2+2·x+5)·(x2+4·x+5) )dx
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