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Pattern - 17 of Indefinite Integration

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

Where, q ( x ) = $\left(a_1\ x-b_1\right)\ \left(a_2\ x-b_2\right)\times\ldots\times\left(a_n\ x-bn\ \right)$

1) I= $\ \frac{x^4-16x^3+10x^2-2x}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\ dx-----\left(1\right)$

Now, ( x – 1 ) ( x – 2 ) ( x – 3 )

= ( $x^2$ – 3x + 2) ( x – 3 )

= $x^3$ – $3x^2$ – $3x^2$ +9x + 2x – 6

= $x^3$ – 6 $x^2$ + 11x – 6

= $x^4-16x^3+10x^2-2x$

= $x^4-6x^3+11x^2-6x-10x^3+60x^2-110x+60-50x^2+108x-60$

$=x\left(x^3-6x^2+11x-6\right)-10\left(x^3-6x^2+11x-6\right)-50x^2+108\ x-60$

$=\ \left(x^3-6x^2+11x-6\right)\times\left(x-10\right)-50x^2+108\ x-60$

$\therefore\ \ I=$ $\int\frac{\left(x^3-6x^2+11x-6\ \right)\left(x-10\right)-50x^2+108x-60\ }{x^3-6x^2+11x-6}dx$

$=\ \int\left[x-10+\ \frac{-50x^2+108x-60}{x^3-6x^2+11x-60}\right]\ dx---\left(2\right)$

Let,

$\frac{-50x^2+108x-60}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\ \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$

∴ -50 $x^2$ + 108x – 60 = A(x-2) (x-3) + B(x-1)(x-3) + C(x-1)(x-2) —-(3)

x = 1 ⇒ -50 + 108 – 60 = A(-1) (-2) + 0 + 0

⇒ -2 = A × 2

⇒ -1 = A

x = 2 ⇒ -200 + 216 – 60 = 0 + B (1) (-1) + 0

⇒ -44 = – B

⇒ B = 44

x = 3 ⇒ -450 + 324 – 60 = 0 + 0 + c (2) (1)

⇒ -186 = 2c

⇒ -93 = c

∴ I = ∫(x – 10) dx + (-1) ∫ $\frac{1}{x-1} dx$ + 44 ∫ $\frac{1}{x-2}$ dx – 93 ∫ $\frac{1}{x-3} dx$

= $\ \frac{x^2}{2}$ – 10x – log | x – 1 | + 44 log | x – 2 | – 93 log | x – 3 | + c