Question

complete the integration ​

Answer 1

Question :-

$\rm \: \displaystyle\int\rm \frac{1}{ {x}^{3} - {x}^{2} - x + 1} \: dx$

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm \frac{1}{ {x}^{3} - {x}^{2} - x + 1} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1}{({x}^{3} - {x}^{2}) - (x - 1)} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1}{ {x}^{2} (x - 1) - (x - 1)} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1}{ ({x}^{2} - 1) (x - 1)} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1}{ (x + 1)(x - 1) (x - 1)} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1}{ (x + 1)(x - 1)^{2} } \: dx$

Now, to integrate this integral, we use method of Partial Fractions.

Let assume that

$\rm \: \frac{1}{ (x + 1)(x - 1)^{2} } = \dfrac{a}{x + 1} + \dfrac{b}{x - 1} + \dfrac{c}{ {(x - 1)}^{2} } - - - (1)$

On taking LCM, we get

$\rm \: a {(x - 1)}^{2} + b(x - 1)(x + 1) + c(x + 1) = 1$

On substituting x = 1, we get

$\rm \: 2c = 1 \rm\implies \:c = \frac{1}{2}$

On substituting x = - 1, we get

$\rm \: a {( - 1 - 1)}^{2} = 1\rm\implies \:a = \frac{1}{4}$

On substituting x = 0, we get

$\rm \: a - b + c = 1$

$\rm \: \frac{1}{2} - b + \frac{1}{4} = 1$

$\rm \: - b + \frac{3}{4} = 1$

$\rm \: - b = 1 - \frac{3}{4}$

$\rm\implies \:\rm \: b = - \frac{1}{4}$

So, on substituting the values of a, b and c, in equation (1), we get

$\rm \: \frac{1}{ (x + 1)(x - 1)^{2} } = \dfrac{1}{4(x + 1)} - \dfrac{1}{4(x - 1)} + \dfrac{1}{ {2(x - 1)}^{2} }$

On integrating both sides w. r. t. x, we get

$\rm \: \displaystyle\int\rm \frac{1}{ (x + 1)(x - 1)^{2} }dx = \displaystyle\int\rm \dfrac{dx}{4(x + 1)} - \displaystyle\int\rm \dfrac{dx}{4(x - 1)} + \displaystyle\int\rm \dfrac{dx}{ {2(x - 1)}^{2} }$

$\rm \: = \: \dfrac{1}{4} log(x + 1) - \dfrac{1}{4} log(x - 1) - \dfrac{1}{2(x - 1)} + c$

Hence,

$\color{green}\rm\implies \:\displaystyle\int\rm \frac{dx}{ {x}^{3} - {x}^{2} - x + 1} \\ \\\color{green} \rm \: = \: \dfrac{1}{4} log(x + 1) - \dfrac{1}{4} log(x - 1) - \dfrac{1}{2(x - 1)} + c$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$