_ ^ dx 4 x^2 + 9 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{4{x^2} + 9}} = } $

A
$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + c$
B
$\frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + c$
✓
$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + c$
D
$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{3x}}{2}} \right) + c$

✓

Answer

Correct option: C.

$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + c$

c
(c)$\int_{}^{} {\frac{{dx}}{{4{x^2} + 9}} = \frac{1}{4}\int_{}^{} {\frac{{dx}}{{{x^2} + {{(32)}^2}}}} } $
$ = \frac{1}{4}.\frac{2}{3}.{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + c = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + c.$

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