Question

1/(x+1)(2x+1) turn into partial fraction​

Answer 1

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Explanation:

$= > \frac{1}{(x + 1)(2x + 1)}$

$= > \frac{1}{(x + 1)(2x + 1)} = \frac{A}{x + 1} + \frac{B}{2x + 1}$

$= > 1 = A(2x + 1) + B(x + 1)$

$put \: x = - 1$

$= > 1 = A(2( - 1) + 1) + b( - 1 + 1)$

$= > 1 = - A+ 0$

$A = - 1$

$put \: x = 0$

$= > 1 = A(2(0) + 1) + B(1)$

$= > 1 = A+ B$

$= > 1 = - 1 + B$

$B = 2$

$\frac{A}{x + 1} + \frac{B}{2x + 1} = ∫\frac{ - 1}{x + 1}dx + ∫\frac{2}{2x + 1} dx$

$= -log |x + 1| + 2log |2x + 1|$

$= 2log |2x + 1| - log |x + 1|$