How do you integrate $\frac{1}{(x^{2} + 1) (x^{2} + 4)}$ $1 / ((x^2 + 1) (x^2 +4))$ using partial fractions?

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Answer 1 · 2016-09-08T14:35:37

$\frac{1}{6} {2 a r c tan x - a r c tan (\frac{x}{2})} + C$ $1/6{2arc tanx-arc tan (x/2)}+C$ .

Let $I = \int \frac{1}{(x^{2} + 1) (x^{2} + 4)} d x$ $I=int1/((x^2+1)(x^2+4))dx$ .

We will use the Method of Partial Fraction to decompose the

integrand $\frac{1}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{(y + 1) (y + 4)}$ $1/((x^2+1)(x^2+4))=1/((y+1)(y+4))$ , where, $y = x^{2}$ $y=x^2$ .

For this, we have to find $A,B in RR$ such that,

$1/((y+1)(y+4))=A/(y+1)+B/(y+4)$ .

Using Heavyside's Cover-up Method, we have,

$A=[1/(y+4)]_(y=-1)=1/3$ .

$B=[1/(y+1)]_(y=-4)=-1/3$ .

$:. 1/((y+1)(y+4))=(1/3)/(y+1)+(-1/3)/(y+4)$ . Since, $y=x^2$ ,

$1/((x^2+1)(x^2+4))=1/3(1/(x^2+1))-1/3(1/(x^2+4))$ . Therefore,

$I=1/3int(1/(x^2+1))dx-1/3int(1/(x^2+2^2))dx$

$=1/3arc tan x-1/3*1/2arc tan (x/2)$ , i.e.,

$I=1/6{2arc tanx-arc tan (x/2)}+C$ .

Enjoy Maths.!

How do you integrate 1 / ((x^2 + 1) (x^2 +4))1(x2+1)(x2+4)1 / ((x^2 + 1) (x^2 +4)) using partial fractions?