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$\int\frac{1}{\left(4+x\right)^2\left(10+3x\right)}dx$

Final answer to the exercise

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Other solved exercises

$-4x+x-\left(-3\right)$

$1-20+1-\frac{198}{1}$

$-198+-276$

$n+n+n+12<23$

$3\left(x^2-4\right)+4x+x^3$

$\left(2x+3x^2\right)+\left(3x+x^2\right)$

$4-8-4+15$
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