If I1=∫Sin−1⁡2x1+x2dx,I2=∫Cos−1⁡1−x21+x2dx,I3=∫Tan−1⁡2x1−x2dx, then I1+I2−I3=

2∫tan−1⁡xdx+2∫tan−1⁡xdx−2∫tan1⁡xdxI1 = ∫sin-1(2x1+x2)dx = ∫2tan-1xdx = 2∫tan-1x.1dx = 2[tan-1x.x-∫11+x2dx] = 2[xtan-1x-12log(1+x2)]+c = 2xtan-1x-log(1+x2)+c I2 = ∫cos-1(1-x21+x2)dx = ∫2tan-1xdx I3 = ∫tan-1(2x1-x2)dx = ∫2tan-1xdx ∴ I1 = I2 = I3 I1+I2-I3 = 2xtan-1x-log(1+x2)+c

If I1=∫Sin−1⁡2x1+x2dx,I2=∫Cos−1⁡1−x21+x2dx,I3=∫Tan−1⁡2x1−x2dx, then I1+I2−I3=

2∫tan−1⁡xdx+2∫tan−1⁡xdx−2∫tan1⁡xdxI1 = ∫sin-1(2x1+x2)dx = ∫2tan-1xdx = 2∫tan-1x.1dx = 2[tan-1x.x-∫11+x2dx] = 2[xtan-1x-12log(1+x2)]+c = 2xtan-1x-log(1+x2)+c I2 = ∫cos-1(1-x21+x2)dx = ∫2tan-1xdx I3 = ∫tan-1(2x1-x2)dx = ∫2tan-1xdx ∴ I1 = I2 = I3 I1+I2-I3 = 2xtan-1x-log(1+x2)+c

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If $I_{1} = \int {Sin}^{- 1} (\frac{2 x}{1 + x^{2}}) dx, I_{2} = \int {Cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) dx, I_{3} = \int {Tan}^{- 1} (\frac{2 x}{1 - x^{2}}) dx, then I_{1} + I_{2} - I_{3} =$

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$2 {xTan}^{- 1} x - \log (1 + x^{2}) + c$

$2 [{xTan}^{- 1} x - \log (1 + x^{2})] + c$

$2 {xTan}^{- 1} x + \log (1 + x^{2}) + c$

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Detailed Solution

$2 \int \tan^{- 1} xdx + 2 \int \tan^{- 1} xdx - 2 \int \tan^{1} xdx$
$I_{1} = \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx = \int 2 \tan^{- 1} xdx = 2 \int \tan^{- 1} x . 1 dx = 2 [\tan^{- 1} x . x - \int \frac{1}{1 + x^{2}} dx] = 2 [{xtan}^{- 1} x - \frac{1}{2} \log (1 + x^{2})] + c = 2 {xtan}^{- 1} x - \log (1 + x^{2}) + c I_{2} = \int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) dx = \int 2 \tan^{- 1} xdx I_{3} = \int \tan^{- 1} (\frac{2 x}{1 - x^{2}}) dx = \int 2 \tan^{- 1} xdx ∴ I_{1} = I_{2} = I_{3} I_{1} + I_{2} - I_{3} = 2 {xtan}^{- 1} x - \log (1 + x^{2}) + c$