Question

1
The partial fractions of 1/(x^2+9)(x^2+16) are​

Answer 1

Q) The partial fractions of $\frac{1}{(x^{2} +9)(x^{2} +16) }$ are.

a) $\frac{1}{7} [\frac{1}{x^{2} +9} - \frac{1}{x^{2} +16}$].

b) $\frac{1}{9} [ \frac{1}{x^{2} +9}- \frac{1}{x^{2} +16}$]

c) $\frac{1}{7}[\frac{1}{x^{2} +16 } - \frac{1}{x^{2} + 9} ]$.

d) $\frac{1}{25}[\frac{1}{x^{2} +9} -\frac{1}{x^{2} +16} ]$

Option (a)$\frac{1}{7} [\frac{1}{x^{2} +9} - \frac{1}{x^{2} +16}$ is the right answer.

Given,

$\frac{1}{(x^{2} +9)(x^{2} +16)}$

To Find,

The partial fractions of $\frac{1}{(x^{2} +9)(x^{2} +16)}$ .

Solution,

Here, given that

the equation $\frac{1}{(x^{2} +9)(x^{2} +16)}$.

It can be written as,$\frac{(x^{2} +16) - (x^{2} +9)}{(x^{2} +9)(x^{2} +16)}$.

The value of the difference in the numerator is 7, so the expression has to be divided by 7.

$[\frac{(x^{2} +16) - (x^{2} +9)}{(x^{2} +9)(x^{2} +16)}] \frac{1}{7}$

so, it can be written as  $\frac{1}{7} [ \frac{1}{x^{2} +9} - \frac{1}{x^{2} +16}]$.

Hence, the partial fractions of $\frac{1}{(x^{2} +9)(x^{2} +16)}$  are $\frac{1}{7} [ \frac{1}{x^{2} +9} - \frac{1}{x^{2} +16} ]$.

#SPJ1

Answer 2

The partial fraction of 1/(x^2+9)(x^2+16) is 1/7[1/x^+9 - 1/x^2+16]

Given:

The equation, 1/(x^2+9)(x^2+16).

To Find:

The partial fractions of 1/(x^2+9)(x^2+16).

Solution:

In the given equation, the denominator has a polynomial degree of 2, and the numerator should be assumed as ax+b and cx+d.

So the equation will be,

1/(x^2+9)(x^2+16) = ax+b/x^2+9 + cx+d/ x^2+16

By taking LCM on R.H.S,

1/(x^2+9)(x^2+16) = (ax+b)(x^2+16) + (cx+d)(x^2+9) /(x^2+9)(x^2+16)

By cancelling the denominators we get,

1 =  (ax+b)(x^2+16) + (cx+d)(x^2+9) ___eqn(1)

Multiplying the coefficients,

1 = Ax^3+16Ax+Bx^2+16B+Cx^2+9Cx+Dx^2+9D

By comparing the coefficients,

x^3=> 0 = A+C

x^2=> 0= B+D

x=> 16A+9C

1=> 16B+9D

By solving these equations,

We get the values of A,B,C,D are;

A=0  B=1/7  C=0  D=-1/7

Substituting the values in eqn(1),

We get,

1/7[1/x^+9 - 1/x^2+16]