Question

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

Accepted Answer

$2\,\,\left[ {\frac{{{{\left( {x + 1} \right)}^{7/2}}}}{7}\,\, - \,\,2\frac{{{{\left( {x + 1} \right)}^{5/2}}}}{5}\,\, + \,\,2\frac{{{{\left( {x + 1} \right)}^{3/2}}}}{3}} \right] + \,\,c$

Answer

$\frac{{{{\left( {x + 1} \right)}^{7/2}}}}{7}\,\, - \,\,2\,\,\frac{{{{\left( {x + 1} \right)}^{5/2}}}}{5}\,\, + \,\,2\frac{{{{\left( {x + 1} \right)}^{3/2}}}}{3}\,\, + \,\,c$

Answer

$\frac{{{{\left( {x + 1} \right)}^{7/2}}}}{7}\,\, - \,\,2\frac{{{{\left( {x + 1} \right)}^{5/2}}}}{5}\,\, + \,\,5$

Answer

$\frac{{{{\left( {x + 1} \right)}^{7/2}}}}{7}\,\, - \,\,3\frac{{{{\left( {x + 1} \right)}^{5/2}}}}{5}\,\, + \,\,11{\left( {x + 1} \right)^{1/2}}\,\, + \,\,c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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