∫ (((√(1+x^2 )) dx)/x^4 ) ?


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Question Number 107036 by john santu last updated on 08/Aug/20

$\int \frac{\sqrt{1 + x^{2}} dx}{x^{4}} ?$
Commented by kaivan.ahmadi last updated on 08/Aug/20

$x = t g u \Rightarrow d x = (1 + {t g}^{2} u) d u = {s e c}^{2} u d u$ $1 + x^{2} = {s e c}^{2} u$ $x^{4} = {t g}^{4} u$ $\Rightarrow \int \frac{s e c u}{{t g}^{4} u} ({s e c}^{2} u) d u = \int \frac{{s e c}^{3} u}{{t g}^{4} u} d u = \int \frac{c o s u}{{s i n}^{4} u} d u$ $n o w l e t v = s i n u \Rightarrow d v = c o s u d u$ $\Rightarrow \int \frac{d v}{v^{4}} = \int v^{- 4} d v = \frac{v^{- 3}}{- 3} + c = \frac{- 1}{3 v^{3}} + c =$ $\frac{- 1}{3 {s i n}^{3} u} + c = \frac{- 1}{3 s i n ({t g}^{- 1} x)} + c$
	Answered by Dwaipayan Shikari last updated on 08/Aug/20

	$\int \frac{\sqrt{1 + \tan^{2} θ}}{\tan^{4} θ} \sec^{2} θ d θ$ $\int \frac{\sec^{3} θ}{\tan^{4} θ} d θ (x = \tan θ 1 = \sec^{2} θ \frac{d θ}{dx}$ $\int \frac{\cos θ}{\sin^{4} θ} d θ$ $- \frac{1}{{3 \sin}^{3} θ} + C = - \frac{{(1 + x^{2})}^{\frac{3}{2}}}{{3 x}^{3}} + C$
	Answered by bobhans last updated on 08/Aug/20

	$\hat{} bobhan \hat{s}$ $\int \frac{\sqrt{x^{2} (1 + \frac{1}{x^{2}})}}{x^{4}} dx = \int \frac{x \sqrt{1 + \frac{1}{x^{2}}} dx}{x^{4}}$ $= \int \frac{\sqrt{1 + \frac{1}{x^{2}}}}{x^{3}} dx . [let u = 1 + \frac{1}{x^{2}}]$ $du = - {2 x}^{- 3} dx \Rightarrow \frac{dx}{x^{3}} = - \frac{du}{2}$ $- \int \frac{u^{1 / 2}}{2} du = - \frac{1}{2} \times \frac{2}{3} u^{3 / 2} + c$ $- \frac{1}{3} \sqrt{{(1 + \frac{1}{x^{2}})}^{3}} + c = - \frac{x^{2} + 1}{{3 x}^{2}} \times \frac{\sqrt{x^{2} + 1}}{x} + c$ $= - \frac{(x^{2} + 1) \sqrt{x^{2} + 1}}{{3 x}^{3}} + c$
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