If ∫ (1/((x^2 +1)(x^2 +4))) dx = A tan^(−1) x+B tan^(−1) (x/2)+C, then


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Question Number 86897 by ram roop sharma last updated on 01/Apr/20

$If \int \frac{1}{(x^{2} + 1) (x^{2} + 4)} d x$ $= A \tan^{- 1} x + B \tan^{- 1} \frac{x}{2} + C, then$
Commented by Tony Lin last updated on 01/Apr/20
$(d/dx)(Atan^(−1) x+Btan^(−1) (x/2)+C) =(A/(x^2 +1))+(B/(((x/2))^2 +1))×(1/2) =(A/(x^2 +1))+((2B)/(x^2 +4)) =((A(x^2 +4)+2B(x^2 +1))/((x^2 +1)(x^2 +4))) ⇒ { ((A+2B=0)),((4A+2B=1)) :} ⇒A=(1/3) , B=−(1/6)$
$\frac{d}{d x} ({A t a n}^{- 1} x + {B t a n}^{- 1} \frac{x}{2} + C)$ $= \frac{A}{x^{2} + 1} + \frac{B}{{(\frac{x}{2})}^{2} + 1} \times \frac{1}{2}$ $= \frac{A}{x^{2} + 1} + \frac{2 B}{x^{2} + 4}$ $= \frac{A (x^{2} + 4) + 2 B (x^{2} + 1)}{(x^{2} + 1) (x^{2} + 4)}$ $\Rightarrow {\begin{cases} A + 2 B = 0 \\ 4 A + 2 B = 1 \end{cases}$ $\Rightarrow A = \frac{1}{3}, B = - \frac{1}{6}$
	Answered by redmiiuser last updated on 01/Apr/20

	$A = \frac{1}{3}$ $B = - \frac{1}{6}$
	Commented by redmiiuser last updated on 01/Apr/20
	$simple way to solve is by partial fractions.$
	$s i m p l e w a y t o s o l v e$ $i s b y p a r t i a l f r a c t i o n s .$
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