Find the value of $\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$ ?

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Find the value of $\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$ ?

$\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$

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Answer 1 · 2017-09-22T17:15:30

See below.

Explanation:

) I used $tan (u + v) = \frac{tan u + tan v}{1 - tan u \cdot tan v}$ $tan(u+v)=(tanu+tanv)/(1-tanu*tanv)$ identity for decomposing $arctan (\frac{2 x - 1}{1 + x - x^{2}})$ $arctan((2x-1)/(1+x-x^2))$ .

2) I used $x = 1 - u$ $x=1-u$ transform in $I$ $I$ integral.

3) After summing 2 integrals, I found result.

$I$ $I$ = $\int_{0}^{1} arctan [\frac{2 x - 1}{1 + x - x^{2}}] \cdot d x$ $int_0^1 arctan[(2x-1)/(1+x-x^2)]*dx$

= $\int_{0}^{1} arctan (\frac{2 x - 1}{1 - x \cdot (x + 1)}) \cdot d x$ $int_0^1 arctan((2x-1)/(1-x*(x+1)))*dx$

= $\int_{0}^{1} arctan x \cdot d x$ $int_0^1 arctanx*dx$ + $\int_{0}^{1} arctan (x - 1) \cdot d x$ $int_0^1 arctan(x-1)*dx$

After using $x = 1 - u$ $x=1-u$ an $d x = - d u$ $dx=-du$ transforms,

$I$ $I$ = $\int_{1}^{0} arctan (1 - u) \cdot (- d u)$ $int_1^0 arctan(1-u)*(-du)$ + $\int_{1}^{0} arctan (- u) \cdot (- d u)$ $int_1^0 arctan(-u)*(-du)$

= $\int_{1}^{0} arctan (u - 1) \cdot d u$ $int_1^0 arctan(u-1)*du$ + $\int_{1}^{0} arctan u \cdot d u$ $int_1^0 arctanu*du$

= $- \int_{0}^{1} arctan u \cdot d u$ $-int_0^1 arctanu*du$ - $\int_{0}^{1} arctan (u - 1) \cdot d u$ $int_0^1 arctan(u-1)*du$

= $- \int_{0}^{1} arctan x \cdot d x$ $-int_0^1 arctanx*dx$ - $\int_{0}^{1} arctan (x - 1) \cdot d x$ $int_0^1 arctan(x-1)*dx$

After summing 2 integrals,

$2 I = 0$ $2I=0$ , hence $I = 0$ $I=0$ .

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Find the value of $\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$ ?

$\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$

1 Answer

Explanation:

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Find the value of $\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$ ?

$\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ $int_0^1tan^-1((2x-1)/(1+x-x^2))dx$