$\mathop \smallint \nolimits_0^3 \,\,\frac{{3x\,\, + \,\,1}}{{{x^2}\,\, + \,\,9}}\,\,dx$ is equal to

$\log \,\,\left( {2\,\,\sqrt 2 } \right)\,\, + \,\,\frac{{\rm{\pi }}}{{12}}$

$\log \,\,\left( {2\,\,\sqrt 2 } \right)\,\, + \,\,\frac{{\rm{\pi }}}{2}$

$\log \,\,\left( {2\,\sqrt 2 } \right)\,\, + \,\,\frac{{\rm{\pi }}}{6}$

$\log \,\,\left( {2\,\sqrt 2 } \right)\,\, + \,\,\frac{{\rm{\pi }}}{3}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Medium

1 of 1

Application +4 | -1

\int \nolimits_0^3 \,\,\frac{{3x\,\, + \,\,1}}{{{x^2}\,\, + \,\,9}}\,\,dx

is equal to

View answer and explanation

Correct option: A

\log \,\,\left( {2\,\,\sqrt 2 } \right)\,\, + \,\,\frac{\mathrm{\pi }}{{12}}

Option A (

\log \,\,\left( {2\,\,\sqrt 2 } \right)\,\, + \,\,\frac{\mathrm{\pi }}{{12}}

) is correct because it is the result of correctly splitting the integral and evaluating each part.