Integrals — Class 12 Maths Solution

Let $I = \int {{{\tan }^{ - 1}}x} dx = \int {{{\tan }^{ - 1}}x \cdot 1} dx$

$= {\tan ^{ - 1}}x\int {\left( 1 \right)dx} - \int {\left( {\cfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) \cdot \int {\left( 1 \right)dx} } \right)dx}$

$= x{\tan ^{ - 1}}x - \cfrac{1}{2}\int {\cfrac{{2x}}{{1 + {x^2}}}dx + C = x{{\tan }^{ - 1}}x - \cfrac{1}{2}\log \left( {1 + {x^2}} \right) + C}$

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