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Pattern - 18 of Indefinite Integration

February 18, 2021
Category: Pattern - 18 of Indefinite Integration

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⋆ Pattern – 18

I= $\ \int\frac{p\left(x\right)}{q\left(x\right)}\ dx_1\$ where, d ( p ( x ) ) < d ( q ( x ) )

Where, q ( x ) = ( $a_1$ x – $b_1$ ) ( $a_2x - b_2$ ) × ( $a_3x - b_3$ ) × ( $a_nx - b_n$ )

1) $\ I=\ \int\frac{x^2}{\left(x-1\right)^3\ \left(x+1\right)}\ dx$ —-(1)

Here,

$\frac{x^2}{\left(x-1\right)^3\ \left(x+1\right)}$ = $\ \frac{A}{x-1}+\ \frac{B}{\left(x-1\right)^2}$ + $\ \frac{C}{\left(x-1\right)^3}$ + $\ \frac{D}{x+1}$ —-(2)

∴ $x^2 = A \left(x-1\right)^2 ( x+1 )$ + B (x – 1)(x + 1) + C(x + 1) + D $\left(x-1\right)^3$ —-(3)

x = 1 ⇒ 1 = c ( 1 + 1 )

⇒ 1 = 2c

⇒ $\frac{1}{2}$ = c

x = -1 ⇒ 1 = D $\left(-2\right)^3$

= -8D

⇒ $\frac{-1}{8}$ = D

We have, c = $\frac{1}{2}$ , D = $\frac{-1}{8}$

x = 2 ⇒ 4 = A $\left(1\right)^2(3)$ + B(1)(3) + $\frac{1}{2}( 2+1)$ – $\frac{1}{8}$ $\left(2-1\right)^3$

⇒ 4 = 3A + 3B + $\frac{3}{2}-\frac{1}{8}$

⇒ 32 = 24A + 24B + 12 – 1

⇒ 32 = 24A + 24B + 11

⇒ 21 = 24A + 24B

Now,

x = 0 ⇒ 0 = A × 1 × 1 + B(-1)(1) + $\frac{1}{2}$ (1) – $\frac{1}{8}$ (-1)

⇒ 0 = A – B + $\frac{1}{2}$ + $\frac{1}{8}$

⇒ 0 = 8A – 8B + 4 + 1

⇒ -5 = 8A – 8B

Hence, we have

24A + 24B = 21 ——(4)

8A – 8B = -5 ———-(5)

from (4) & (5)

24A + 24B = 21

24A – 24B = -15

48A = 6

∴ A = $\frac{1}{8}$

from (5),

1 – 8B = -5

∴ -8B = -6

∴ B = $\frac{3}{4}$

form (3),

$x^2$ = A( $x^2$ – 2x + 1)(x+1) + B ( $x^2$ -1) + c ( x + 1) + D( $x^3$ – 3 $x^2$ + 3x – 1)

= A( $x^3 + x^2 - 2x^2$ – 2x + x + 1) + B ( $x^2$ -1 )+c ( x + 1 ) + D( $x^3 -3x^2$ + 3x -1 )

∴ $x^2$ = (A+D) $x^3$ + (-A + B – 3D) $x^2$ + (-A + C + 3D)x + ( A – B + C – D )

Comparing,

A + D = 0

∴ A – $\frac{1}{8}$ = 0

∴ A = $\frac{1}{8}$

-A + B – 3D = 1

∴ $\frac{-1}{8}$ + B + $\frac{3}{8}$ = 1

∴ $\frac{2}{8}$ + B = 1

∴ B = 1 – $\frac{2}{8}$

∴ B = $\frac{6}{8}$

∴ B = $\frac{3}{4}$

-A + C + 3D = 0 &

A – B + C – D = 0

Also, we have, C = $\frac{1}{2}$ , D = $\frac{-1}{8}$ , A = $\frac{1}{8}$ , B = $\frac{3}{4}$

$\therefore I=\ \frac{1}{8}\int\frac{1}{x-1}\ dx$ + $\frac{3}{4}\ \int{{(x-1)}^{-2}\ }dx$ + $\frac{1}{2}\int\left(x-1\right)^{-3}dx$ – $\frac{1}{8}\int\frac{1}{x+1}dx$

= $\frac{1}{8}\log{\left|x-1\right|}$ + $\ \frac{3}{4}\ \frac{\left(x-1\right)^{-1}}{-1}$ + $\ \frac{1}{2}\ \ \frac{\left(x-1\right)^{-2}}{-2}$ – $\frac{1}{8}\log{\left|x+1\right|}+c$