I-F of 1-x2dydx-y-etan-1x is-

The correct option is A e^tan^ 1x dydx+y1+x^2=e^tan^ 1x/_1+x^2 I.F=e^∫ P(x)dx=e^∫11+x^2dx=e^tan^ 1x ...

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Integrating Factor

I.F of 1+x2...

Question

I.F of $(1 + x^{2}) \frac{d y}{d x} + y = e^{{tan}^{- 1} x}$ is:

{tan}^{- 1} x

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e^{{tan}^{- 1} x}

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\frac{1}{1 + x^{2}}

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tan x

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Solution

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Similar questions

\int \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} [{({sec}^{- 1} \sqrt{1 + x^{2}})}^{2} + {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] d x

(x > 0),

^{'} C^{'}

\int \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} . d x

\int e^{{tan}^{- 1} x} [\frac{1}{1 + x^{2}}] d x =

\int e^{t a n^{- 1} x} [\frac{1 + x + x^{2}}{1 + x^{2}}] d x =

\frac{d y}{d x} + \frac{y}{1 + x^{2}} = \frac{e^{t a n^{- 1} x}}{1 + x^{2}}

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