The partial-fraction decomposition of the following expression x2-1xx-13 is kx-1 - px-13 - qx- Find the value of k - p-q-

Let x^2+1x(x 1)^3=Ax+B(x 1)+C(x 1)^2+D(x 1)^3 ....(1)Multiply x(x 1)^3 on both the sides, we get x^2+1=A(x 1)^3+Bx(x 1)^2+Cx(x 1)+D.x ....(2)Pu ...

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Standard XII

Mathematics

Quadratic Polynomial

The partial-f...

Question

The partial-fraction decomposition of the following expression $\frac{x^{2} + 1}{x (x - 1)^{3}}$ is $\frac{k}{x - 1} + \frac{p}{(x - 1)^{3}} + \frac{q}{x}$ . Find the value of $k + p + q$ .

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Solution

Let $\frac{x^{2} + 1}{x (x - 1)^{3}} = \frac{A}{x} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^{2}} + \frac{D}{(x - 1)^{3}}$ ....(1)
Multiply $x (x - 1)^{3}$ on both the sides, we get
$x^{2} + 1 = A (x - 1)^{3} + B x (x - 1)^{2} + C x (x - 1) + D . x$ ....(2)
Put $x = 0$ to find the value of A
$∴ 1 = A (- 1) \Rightarrow A = - 1$
Now put $x = 1$ to find value of D
$∴ D = 2$
Substitute these values in equation (2), we get
$x^{2} + 1 = - 1 (x - 1)^{3} + B x (x - 1)^{2} + C x (x - 1) + 2 x$
Solving this, we get
$x^{2} + 1 = x^{3} (B - 1) + x^{2} (3 - 2 B + c) + x (- 3 + B - C + 2) + 1$
Comparing like coefficients, we get
$B - 1 = 0, 3 - 2 B + c = 1, B - C - 1 = 0$
On solving, we get
$B = 1, C = 0$
Substituting the values of A,B,C and D, we get
$\frac{x^{2} + 1}{x (x - 1)^{3}} = \frac{- 1}{x} + \frac{1}{(x - 1)} + \frac{2}{(x - 1)^{3}}$
Comparing this with $\frac{k}{x - 1} + \frac{p}{(x - 1)^{3}} + \frac{q}{x}$ , we get
$k = 1, p = 2, q = - 1$
Therefore, $k + p + q = 1 + 2 + (- 1) = 2$

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The partial-fraction decomposition of the following expression x2+1x(x−1)3 is kx−1+p(x−1)3+qx. Find the value of k+p+q.

The partial-fraction decomposition of the following expression $\frac{x^{2} + 1}{x (x - 1)^{3}}$ is $\frac{k}{x - 1} + \frac{p}{(x - 1)^{3}} + \frac{q}{x}$ . Find the value of $k + p + q$ .