Let I = ∫√(1+x^2)/x^4dx = ∫√(1+x^2)/x.1/x^3dx =∫√(1+x^2/x^2). 1/x^3dx=∫√(1/x^2+1).1/x^3dx Put 1+1/x^2=t^2⇒ 2/x^3dx=2tdt ⇒ 1/x^3=tdt ∴ I = ∫ t^2dt= t^3 ...

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Byju's Answer

Standard XII

Mathematics

Binomial Integral

∫√1+x2/x4 d x

Question

$\int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x$

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Solution

Let $I = \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x = \int \frac{\sqrt{1 + x^{2}}}{x} . \frac{1}{x^{3}} d x = \int \sqrt{\frac{1 + x^{2}}{x^{2}}} . \frac{1}{x^{3}} d x = \int \sqrt{\frac{1}{x^{2}} + 1} . \frac{1}{x^{3}} d x P u t 1 + \frac{1}{x^{2}} = t^{2} \Rightarrow \frac{- 2}{x^{3}} d x = 2 t d t \Rightarrow - \frac{1}{x^{3}} = t d t ∴ I = - \int t^{2} d t = - \frac{t^{3}}{3} + C = - \frac{1}{3} {(1 + \frac{1}{x^{2}})}^{3 / 2} + C$

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∫√1+x2x4dx

Let I=∫√1+x2x4dx=∫√1+x2x.1x3dx=∫√1+x2x2.1x3dx=∫√1x2+1.1x3dxPut 1+1x2=t2⇒−2x3dx=2tdt⇒−1x3=tdt∴ I=−∫t2dt=−t33+C=−13(1+1x2)3/2+C

$\int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x$