Question

The value of $\smallint \,{e^x}\,\,\frac{{1\,\, + \,\,n{x^{n - 1}}\,\, - \,\,{x^{2n}}}}{{\left( {1\,\, - \,\,{x^n}} \right)\,\,\sqrt {1\,\, - \,\,{x^{2n}}} }}\,\,dx$ is equal to

Accepted Answer

$\frac{{{e^x}\,\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}\,\, + \,\,c$

Answer

${e^x}\,\,\left( {\sqrt {1 - {x^2}} } \right)\,\, + \,\,c$

Answer

${e^x}\,\,\frac{{\sqrt {1 + {x^{2n}}} }}{{1 + {x^{2n}}}}\,\, + \,\,c$

Answer

$\frac{{{e^x}\,\sqrt {1 - {x^n}} }}{{1 - {x^{2n}}}}\,\, + \,\,c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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