Question

________________INTEGRATE !!​

Answer 1

$\int _0^1\:tan^{-1}\left(\frac{1}{x^2-x+1}\right)dx$

I used  

tan

(

u

+

v

)

=

tan

u

+

tan

v

1

−

tan

u

⋅

tan

v

identity for decomposing  

arctan

(

2

x

−

1

1

+

x

−

x

2

)

.

2) I used  

x

=

1

−

u

transform in  

I

integral.

3) After summing 2 integrals, I found result.

I

=

∫

1

0

arctan

[

2

x

−

1

1

+

x

−

x

2

]

⋅

d

x

=

∫

1

0

arctan

(

2

x

−

1

1

−

x

⋅

(

x

+

1

)

)

⋅

d

x

=

∫

1

0

arctan

x

⋅

d

x

+

∫

1

0

arctan

(

x

−

1

)

⋅

d

x

After using  

x

=

1

−

u

an  

d

x

=

−

d

u

transforms,

I

=

∫

0

1

arctan

(

1

−

u

)

⋅

(

−

d

u

)

+

∫

0

1

arctan

(

−

u

)

⋅

(

−

d

u

)

=

∫

0

1

arctan

(

u

−

1

)

⋅

d

u

+

∫

0

1

arctan

u

⋅

d

u

=

−

∫

1

0

arctan

u

⋅

d

u

-

∫

1

0

arctan

(

u

−

1

)

⋅

d

u

=

−

∫

1

0

arctan

x

⋅

d

x

-

∫

1

0

arctan

(

x

−

1

)

⋅

d

x

After summing 2 integrals,

2

I

=

0

, hence  

I

=

0

.

Answer 2

Step-by-step explanation:

$\sf \: Let, \frac{\mathrm{d}}{\mathrm{d}x}f_{x} = \frac{\mathrm{d} }{\mathrm{d} x}tan^{-1}\frac{\sqrt(1 + x^{2}) + \sqrt(1 - x^{2})}{\sqrt(1 + x^{2})- \sqrt(1 - x^{2})} \\ \\ </p><p> </p><p></p><p> \sf \: f_{x} = tan^{-1}\frac{\sqrt(1 + x^{2}) + \sqrt(1 - x^{2})}{\sqrt(1 + x^{2})- \sqrt(1 - x^{2})}</p><p> \\ \\ </p><p> \sf \: f_{x} = tan^{-1}(\frac{2\sin x}{2\cos x}) = tan^{-1}(\tan x) </p><p>$