How do you integrate $(x^3+4)/(x^2+4)$ using partial fractions?

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Answer 1 · 2016-09-02T10:13:02

$x^2/2-2ln(x^2+4)+2arc tan (x/2)+C$ .

Let $I=int(x^3+4)/(x^2+4)dx$

Since the degree of poly. in Nr. $>$ that of in DR. , we have

to first perform long division. Instead, we proceed as under :

$Nr.=x^3+4$

$=x^3+4x-4x+4=x(x^2+4)-4x+4$ . Hence,

$(x^3+4)/(x^2+4)={x(x^2+4)-4x+4}/(x^2+4)$

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

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

$:. I=int[x-(4x)/(x^2+4)+4/(x^2+4)]dx$

$=intxdx-2int(2x)/(x^2+4)dx+4int1/(x^2+4)dx$

$=x^2/2-2int(d(x^2+4))/(x^2+4)+4*1/2*arc tan (x/2)$

$=x^2/2-2ln(x^2+4)+2arc tan (x/2)+C$ .

Enjoy maths.!

How do you integrate (x^3+4)/(x^2+4)(x^3+4)/(x^2+4) using partial fractions?