Question

$\smallint \,\,\frac{{\log \,\,\left( {x\,\, + \,\,\sqrt {1 + {x^2}} } \right)}}{{\sqrt {1\,\, + \,\,{x^2}} }}\;\,\,dx$ is equal to

Accepted Answer

$\frac{1}{2}\,\,{\left[ {\log \,\,\left( {x\,\, + \,\,\sqrt {1\,\, + \,\,{x^2}} } \right)} \right]^2} + \,\,c$

Answer

${\left[ {\,\log \,\,\left( {x\,\, + \,\,\sqrt {1\,\, + \,\,{x^2}} } \right)\,} \right]^2} + \,\,c$

Answer

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

Answer

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

More Questions from INDEFINITE INTEGRATION