Question

plz solve this question​

Answer 1

Step-by-step explanation:

hope it helps.......✌✌

Answer 2

Answer:

$\frac{tan^{-1} (\sqrt{2x} +1) + tan^{-1} (\sqrt{2x}-1)}{\sqrt{2} } + C$

Step-by-step explanation:

Factor the denominator and apply partial fractions

$\int\limit { \frac{x^2+1}{x^4+1}  } \, dx = \int\limit { \frac{x^2+1}{(x^2+\sqrt{2}x +1)(x^2-\sqrt{2}x +1)}  } \, dx\\\int\limit { \frac{1}{2(x^2+\sqrt{2}x +1)}  } \, dx + \int\limit { \frac{1}{2(x^2-\sqrt{2}x +1)}  } \, dx$

Now use complete square method

$x^2+\sqrt{2}x +1 = (x + \frac{1}{\sqrt{2} } )^2 + \frac{1}{2}$

$x^2-\sqrt{2}x +1 = (x - \frac{1}{\sqrt{2} } )^2 + \frac{1}{2}$

so we get

$\int\limit { \frac{1}{2((x + \frac{1}{\sqrt{2} } )^2 + \frac{1}{2})}  } \, dx + \int\limit { \frac{1}{2((x - \frac{1}{\sqrt{2} } )^2 + \frac{1}{2})}  } \, dx$

Now take

$u = \sqrt{2}x + 1, \frac{du}{dx} = \sqrt{2}\\ v = \sqrt{2}x - 1, \frac{dv}{dx} = \sqrt{2}$

so we have

$\int\limit { \frac{1}{2((x + \frac{1}{\sqrt{2} } )^2 + \frac{1}{2})}  } \, dx + \int\limit { \frac{1}{2((x - \frac{1}{\sqrt{2} } )^2 + \frac{1}{2})}  } \, dx =\\\\\int\limits { \frac{\sqrt{2} }{ 2(u^2+1) }  } \, du + \int\limits { \frac{\sqrt{2} }{ 2(v^2+1) }  } \, dv\\= \frac{ tan^{-1} (u) + tan^{-1} (v) }{ \sqrt{2}  } + C \\ = \frac{ tan^{-1} (\sqrt{2}x + 1) + tan^{-1} (\sqrt{2}x - 1) }{ \sqrt{2}  } + C$