Compute the integral of f(x)=x^3/x^4+1

A basic function of integration states that: for a function f(x), the integral of f'(x)/f(x) = ln[f(x)] (the natural log of the modulus of f(x)). Take the denominator of f(x), x4+1. We will refer to this as j(x) Differentiating this denominator gives : 4x= j'(x) Therefore, the numerator, x= 1/4j'(x) Having estabilished this, we can rewrite the integral of f(x) as such : integral ( 0.25j'(x)/j(x)) dx Taking the constant value, 1/4, out of the integral, we are left with: integral( j'(x)/j(x)) dx Above, we have estabilished that the integral of f'(x) / f(x) is ln[f(x)]. Therefore, if we substitue j(x) into this result, we are left with: ln(x4+1). However, this is not yet the final answer! We must remember to reinsert the constant we took out of the integral: 1/4. We also have the unknown constant to add, c, which is added after any integration. Therefore, the final answer is 1/4ln(x4+1) + c

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