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MHT CET 2017 Math Solved Paper

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Question : 26 of 50

Marks: +1, -0

If

\int \frac{1}{(x^{2}+4)(x^{2}+9)} \, dx

A \tan^{-1} \frac{x}{2}

B \tan^{-1} \left( \frac{x}{3} \right)

+ C, then A - B =

$\frac{1}{6}$
$\frac{1}{30}$
$-\frac{1}{30}$
$-\frac{1}{6}$

Solution:

(a) Consider

I = \int \frac{1}{(x^{2}+4)(x^{2}+9)} \, dx

= \int \frac{1}{5} \left( \frac{1}{x^{2}+4} - \frac{1}{x^{2}+9} \right) \, dx

= \frac{1}{5} \left[ \int \frac{1}{x^{2}+2^{2}} \, dx - \int \frac{1}{x^{2}+3^{2}} \, dx \right]

= \frac{1}{5} \left[ \frac{1}{2} \tan^{-1} \frac{x}{2} - \frac{1}{3} \tan^{-1} \frac{x}{3} \right]

+C

= \left[ \frac{1}{10} \tan^{-1} \frac{x}{2} - \frac{1}{15} \tan^{-1} \frac{x}{3} \right]

+ C

We have,

I = A \tan^{-1} \frac{x}{2}

+

B \tan^{-1} \frac{x}{3}

+ C

So,

A \tan^{-1} \frac{x}{2}

+

B \tan^{-1} \frac{x}{3}

+ C =

\frac{1}{10} \tan^{-1} \frac{x}{2} - \frac{1}{15} \tan^{-1} \frac{x}{3}

+ C

After comparing on both sides, we get

A =

\frac{1}{10}

and B =

\frac{1}{15}

Hence, A - B =

\frac{1}{10} - \frac{1}{15}

=

\frac{10+15}{150} = \frac{1}{6}

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