Question

If $I = \smallint \,\,\frac{1}{{{x^4}\,\,\sqrt {{a^2}\,\, + \,\,{x^2}} }}\,\,dx$, then I equals

Accepted Answer

$\frac{1}{{{a^4}}}\,\,\left\{ {\frac{1}{x}\,\,\sqrt {{a^2}\,\, + \,\,{x^2}} \,\, - \,\,\frac{1}{{3{x^3}}}\,\,{{\left( {{a^2}\,\, + \,\,{x^2}} \right)}^{3/2}}} \right\} + C$

Answer

$\frac{1}{{{a^4}}}\,\,\left\{ {\frac{1}{x}\,\,\sqrt {{a^2}\,\, + \,\,{x^2}} \,\, - \,\,\frac{1}{{3{x^3}}}\,\,\sqrt {{a^2}\,\, + \,\,{x^2}} } \right\} + C$

Answer

$\frac{1}{{{a^4}}}\,\,\left\{ {\frac{1}{x}\sqrt {{a^2}\,\, + \,\,{x^2}} - \,\,\frac{1}{{2\,\sqrt x }}{{\left( {{a^2}\,\, + \,\,{x^2}} \right)}^{3/2}}} \right\} + C$

Answer

None of these

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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