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Pattern - 19 of Indefinite Integration

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⋆ Pattern – 19

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

Where, q ( x ) = (a_1x – b_1) (a_2x - b_2) × (a_3x - b_3) × (a_nx - b_n )

1)\ I=\int\frac{5x}{\left(x+1\right)\left(x^2+9\right)}dx—(1)

\frac{5x}{\left(x+1\right)\left(x^2+9\right)} = \frac{A}{x+1}+\ \frac{Bx+c}{x^2+9}—-(2)

∴  5x = A(x^2+9) + (Bx + c) (x + 1) ——(3)

x = -1 ⇒ -5 = A(10)

⇒ – \frac{1}{2} = A

x = 0  ⇒ 0 = \frac{-1}{2} (9) + c(1)

⇒ \frac{9}{2} = c

x = 1  ⇒ 5 = \frac{-1}{2} (10) + \left(B+\frac{9}{2}\right)       (2)

⇒ 5 = -5 + 2B + 9

⇒ 1 + 2B

⇒ \frac{1}{2} = B

OR

from (3)

5x = A(x^2+ 9) + x^2B+ Bx + Cx + C

∴  5x = (A + B)x^2 + (B + C) x + (9A + C)

Comparing both sides, we have

∴  A + B = 0

∴  \frac{-1}{2} + B = 0

∴  B = \frac{1}{2}

 B + C = 5 9A + C = 0

∴  C = -9A

= \frac{9}{2}

A = \frac{-1}{2}, B = \frac{1}{2}, C = \frac{9}{2}

\therefore\ I = \ \frac{-1}{2}\int\frac{1}{x+1}dx + \frac{1}{2}\int\frac{x+9}{x^2+9}dx

=\ \frac{-1}{2}\int\frac{1}{x+1}dx +\ \frac{1}{2}\times\ \frac{1}{2}\ \int\frac{2x}{x^2+9}\ dx+\frac{9}{2}\int\frac{1}{x^2+9}\ dx

=\ \frac{-1}{2}\ \int\frac{1}{x+1}dx + \ \frac{1}{4}\int\frac{2x}{x^2+9}dx + \frac{9}{2}\int\frac{1}{x^2+9}dx

= \ \frac{-1}{2}\log{\left|x+1\right|} + \frac{1}{4}\log{\left|x^2+9\right|} + \frac{9}{2}\times\frac{1}{3}\tan^{-1}{\frac{x}{3}}+C

 

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