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

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Answer 1 · 2016-09-02T10:13:02

$\frac{x^{2}}{2} - 2 ln (x^{2} + 4) + 2 a r c tan (\frac{x}{2}) + C$ $x^2/2-2ln(x^2+4)+2arc tan (x/2)+C$ .

Let $I = \int \frac{x^{3} + 4}{x^{2} + 4} d x$ $I=int(x^3+4)/(x^2+4)dx$

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

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

$N r . = x^{3} + 4$ $Nr.=x^3+4$

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

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

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

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

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

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

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

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

Enjoy maths.!

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