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Question Number 84708 by Power last updated on 15/Mar/20

	Answered by TANMAY PANACEA last updated on 16/Mar/20
	$∫(dx/((x+2)^2 (√(x^2 +2x−5)))) ∫(dx/((x+2)^2 (√((x+1)^2 −6)))) x+2=(1/t)→dx=((−dt)/t^2 ) ∫((−dt)/(t^2 ×(1/t^2 )(√(((1/t)−1)^2 −6)))) ∫((−dt)/(√((1/t^2 )−(2/t)−5))) ∫((−t dt)/(√(1−2t−5t^2 ))) (1/(10))∫((−10t−2+2)/(√(−5t^2 −2t+1)))dt now... −5t^2 −2t+1 5{(((√6)/5))^2 −(t+(1/5))^2 } so (1/(10))∫((d(−5t^2 −2t+1))/(√(−5t^2 −2t+1)))+(1/5)∫(dt/((√5) (√((((√6)/5))^2 −(t+(1/5))^2 )) )) (1/(10))×(((−5t^2 −2t+1)^(1/2) )/(1/2))+(1/(5(√5)))sin^(−1) (((t+(1/5))/((√6)/5)))+C pls replace t by (1/(x+2))$
	$\int \frac{d x}{{(x + 2)}^{2} \sqrt{x^{2} + 2 x - 5}}$ $\int \frac{d x}{{(x + 2)}^{2} \sqrt{{(x + 1)}^{2} - 6}}$ $x + 2 = \frac{1}{t} \to d x = \frac{- d t}{t^{2}}$ $\int \frac{- d t}{t^{2} \times \frac{1}{t^{2}} \sqrt{{(\frac{1}{t} - 1)}^{2} - 6}}$ $\int \frac{- d t}{\sqrt{\frac{1}{t^{2}} - \frac{2}{t} - 5}}$ $\int \frac{- t d t}{\sqrt{1 - 2 t - 5 t^{2}}}$ $\frac{1}{10} \int \frac{- 10 t - 2 + 2}{\sqrt{- 5 t^{2} - 2 t + 1}} d t$ $n o w . . .$ $- 5 t^{2} - 2 t + 1$ $5 {{(\frac{\sqrt{6}}{5})}^{2} - {(t + \frac{1}{5})}^{2}}$ $s o$ $\frac{1}{10} \int \frac{d (- 5 t^{2} - 2 t + 1)}{\sqrt{- 5 t^{2} - 2 t + 1}} + \frac{1}{5} \int \frac{d t}{\sqrt{5} \sqrt{{(\frac{\sqrt{6}}{5})}^{2} - {(t + \frac{1}{5})}^{2}}}$ $\frac{1}{10} \times \frac{{(- 5 t^{2} - 2 t + 1)}^{\frac{1}{2}}}{\frac{1}{2}} + \frac{1}{5 \sqrt{5}} {s i n}^{- 1} (\frac{t + \frac{1}{5}}{\frac{\sqrt{6}}{5}}) + C$ $p l s r e p l a c e t b y \frac{1}{x + 2}$
	Commented by Power last updated on 15/Mar/20

	$thanks$
	Commented by TANMAY PANACEA last updated on 15/Mar/20

	$m o s t w e l c o m e$
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