How do you find the indefinite integral of $\int \frac{x^{4} + x - 4}{x^{2} + 2}$ $int (x^4+x-4)/(x^2+2)$ ?

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Answer 1 · 2018-04-13T07:50:33

The answer is $= \frac{x^{3}}{3} - 2 x + \frac{1}{2} ln (x^{2} + 2) + C$ $=x^3/3-2x+1/2ln(x^2+2)+C$

We need

$int(u'(x)dx)/(u(x))=ln(u(x))+C$

Perform a polynomial long division

$(x^4+x-4)/(x^2+2)=x^2-2+(x)/(x^2+2)$

Therefore,

$int((x^4+x-4)dx)/(x^2+2)=intx^2dx-int2dx+1/2int(2xdx)/(x^2+2)$

$=x^3/3-2x+1/2ln(x^2+2)+C$

How do you find the indefinite integral of int (x^4+x-4)/(x^2+2)∫x4+x−4x2+2int (x^4+x-4)/(x^2+2)?