Mathematics

Solve $$\int\limits_1^{-1} {\dfrac{d}{{dx}}ta{n^{ - 1}}\left( {\dfrac{1}{x}} \right)dx} $$


SOLUTION

Consider the given integral.

$$I=\int_{1}^{-1}{\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right)}dx$$

$$ I=\int_{1}^{-1}{\dfrac{1}{1+\dfrac{1}{{{x}^{2}}}}}dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \dfrac{d}{dx}\left( {{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}} \right) \right] $$

$$ I=\int_{1}^{-1}{\dfrac{{{x}^{2}}}{{{x}^{2}}+1}}dx $$

$$ I=\int_{1}^{-1}{\dfrac{{{x}^{2}}+1-1}{{{x}^{2}}+1}}dx $$

$$ I=\int_{1}^{-1}{1dx}-\int_{1}^{-1}{\dfrac{1}{{{x}^{2}}+1}}dx $$

$$ I=\left[ x \right]_{1}^{-1}-\left[ {{\tan }^{-1}}\left( x \right) \right]_{1}^{-1} $$

$$ I=\left[ -1-\left( 1 \right) \right]-\left[ {{\tan }^{-1}}\left( -1 \right)-{{\tan }^{-1}}\left( 1 \right) \right] $$

$$ I=\left[ -2 \right]-\left[ {{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)-{{\tan }^{-1}}\left( \dfrac{\pi }{4} \right) \right] $$

$$ I=\left[ -2 \right]-\left[ \dfrac{3\pi }{4}-\dfrac{\pi }{4} \right] $$

$$ I=\left[ -2 \right]-\left[ \dfrac{\pi }{2} \right] $$

$$ I=-2-\dfrac{\pi }{2} $$

 

Hence, this is the answer.

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