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The value of \int \frac{e^{x}\left ( 2-x^{2} \right )dx}{\left ( 1-x \right )\sqrt{1-x^{2}}} is equal to

  • Option 1)

    e^{x}\sqrt{\frac{1+x}{1-x}}+c

  • Option 2)

    e^{2}\sqrt{1+x}+c

  • Option 3)

    e^{2}\sqrt{1-x}+c

  • Option 4)

    e^{x}\sqrt{\frac{1-x}{1+x}}+c

 

Answers (1)

best_answer

As we learnt

Rule for integration by parts -

Take Ist function (u) as according I L A T E

  

- wherein

Where ,

I : Inverse

L : Logarithmic

A : Algebraic 

T : Trignometric

E : Exponential

 

 \int {\frac{{{e^x}\left( {2 - {x^2}} \right)}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} \,\,dx = \int {\frac{{{e^x}\left( {1 - {x^2} + 1} \right)}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} \,dx = \int {\frac{{{e^x}\left( {1 - {x^2}} \right)}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} dx + \int {\frac{{{e^x}}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} \,dx\,\,$=\int {\underbrace {{e^x}\sqrt {\frac{{1 + x}}{{1 - x}}} }_{\pi \,\,\,\,\,\,\,\,\,\,\,\,\,I}} \,dx\, + \int {\frac{{{e^x}}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} \,dx$

Integrate by parts (only first part)

= \sqrt {\frac{{1 + x}}{{1 - x}}} .{e^x} - \int {\frac{1}{2}{{\left( {\frac{{1 + x}}{{1 - x}}} \right)}^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right. \kern-\nulldelimiterspace} 2}}}} \,.\,\frac{d}{{dx}}\left( {\frac{{1 + x}}{{1 - x}}} \right).{e^x}\,dx + \int {\frac{{{e^x}dx}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} \,$

={e^x}.\sqrt {\frac{{1 + x}}{{1 - x}}} - \frac{1}{2}\int {\sqrt {\frac{{1 - x}}{{1 + x}}} } \,.\frac{{1 - x + 1 + x}}{{{{\left( {1 - x} \right)}^2}}}\,{e^x}\,dx + \int {\frac{{{e^x}}}{{\left( {1 - x} \right)\sqrt {1 - {x^2}} }}} \,dx\,$

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


Option 1)

e^{x}\sqrt{\frac{1+x}{1-x}}+c

Option 2)

e^{2}\sqrt{1+x}+c

Option 3)

e^{2}\sqrt{1-x}+c

Option 4)

e^{x}\sqrt{\frac{1-x}{1+x}}+c

Posted by

gaurav

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