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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \log \large\frac{[x +\sqrt {1+x^2}]}{\sqrt {1-x^2}}$$dx$

$(a)\;\frac{1}{2} \log (x +\sqrt {1+x^2})+c \qquad(b)\;\frac{1}{2} \log (x -\sqrt {1+x^2})+c \qquad(c)\;\frac{1}{2} \log (x +\sqrt {1-x^2})+c \qquad (d)\;None$

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1 Answer

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$ \log [x + \sqrt {1+x^2}]=t$
differentiate with respect to x
$\large\frac{1}{x+\sqrt {1+x^2}} \bigg \{1+ \large\frac{1}{2} \times \frac{2x}{\sqrt {1+x^2}}\bigg\}$$ dx=dt$
$ \large\frac{1}{\sqrt {1+x^2}}$$dx=dt$
=> $\int t.dt$
=> $\large\frac{t^2}{2}$$+c$
$\large\frac{1}{2}$$\log (x +\sqrt {1+x^2})+c$
Hence a is the correct answer.
answered Dec 24, 2013 by meena.p
 
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