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

Question

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

Isabella Ackerman · Accepted Answer

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

Let #I=int1/((x^2+1)(x^2+4))dx#. We will use the Method of Partial Fraction to decompose the integrand #1/((x^2+1)(x^2+4))=1/((y+1)(y+4))#, where, #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.!

Elijah Abram · Answer

undefined To integrate $ \frac{1}{(x^2 + 1)(x^2 + 4)} $ using partial fractions, we first decompose it into partial fractions of the form:

$$ \frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{A}{x^2 + 1} + \frac{B}{x^2 + 4} $$

where $ A $ and $ B $ are constants to be determined. Then, we solve for $ A $ and $ B $ by finding a common denominator:

$$ 1 = A(x^2 + 4) + B(x^2 + 1) $$

Equating coefficients of like terms:

$$ 0x^3 + 0x^2 + 1 = (A + B)x^2 + (4A + B)x + 4A $$

We equate coefficients:

$$ A + B = 0 $$
$$ 4A + B = 0 $$
$$ 4A = 1 $$

Solving these equations yields:

$$ A = \frac{1}{4} $$
$$ B = -\frac{1}{4} $$

Thus, we can write the original expression as:

$$ \frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{1/4}{x^2 + 1} - \frac{1/4}{x^2 + 4} $$

Now, we can integrate each term separately:

$$ \int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx = \frac{1}{4} \int \frac{1}{x^2 + 1} \, dx - \frac{1}{4} \int \frac{1}{x^2 + 4} \, dx $$

These integrals are standard forms and can be evaluated as follows:

$$ \frac{1}{4} \int \frac{1}{x^2 + 1} \, dx = \frac{1}{4} \arctan(x) + C_1 $$
$$ \frac{1}{4} \int \frac{1}{x^2 + 4} \, dx = \frac{1}{8} \ln|x^2 + 4| + C_2 $$

Therefore, the integral of $ \frac{1}{(x^2 + 1)(x^2 + 4)} $ using partial fractions is:

$$ \frac{1}{4} \arctan(x) - \frac{1}{8} \ln|x^2 + 4| + C $$