Q84 / 150 · MHT CET 2018 · Mathematics

If $\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (B x) + C$ then $A + B =$

A
$\frac{9}{4}$
B
$\frac{19}{4}$
C
$\frac{3}{4}$
Correct answer
D
$\frac{13}{12}$

Solution

I = \int \frac{d x}{\sqrt{4^{2} - {(3 x)}^{2}}}

I = \frac{1}{3} \sin^{- 1} (\frac{3 x}{4}) + C

A + B = \frac{1}{3} + \frac{3}{4} = \frac{4 + 9}{12} = \frac{13}{12}

Hence, the correct answer is option: 'D'.

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