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  4. The value of ∫(ex(2-x2)dx/(1-x)√1-x2) is equal to:

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Q. The value of $ \int{\frac{{{e}^{x}}(2-{{x}^{2}})dx}{(1-x)\sqrt{1-{{x}^{2}}}}} $ is equal to:

KEAMKEAM 2001

Solution:

Let $ I=\int{\frac{{{e}^{x}}(2-{{x}^{2}})dx}{(1-x)\sqrt{1-{{x}^{2}}}}} $ $ =\int{\frac{{{e}^{x}}(1-{{x}^{2}})}{(1-x)\sqrt{1-{{x}^{2}}}}dx}+\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx $ $ =\int{{{e}^{x}}}\sqrt{\frac{1+x}{1-x}}dx+\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx $ $ =\left[ {{e}^{x}}\sqrt{\frac{1+x}{1-x}}-\frac{1}{2}\int{\frac{d}{dx}}\sqrt{\frac{1+x}{1-x}}{{e}^{x}}dx \right] $ $ +\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}dx} $ $ ={{e}^{x}}\sqrt{\frac{1+x}{1-x}}-\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx $ $ +\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx $ $ ={{e}^{x}}\sqrt{\frac{1+x}{1-x}}+c $