Q68 / 150 · MHT CET 2016 · Mathematics

$\int (\frac{4 e^{x} - 25}{2 e^{x} - 5}) d x = A x + B \log | 2 e^{x} - 5 | + c$ then

A
$A = 5, B = 3$
Correct answer
B
$A = 5, B = - 3$
C
$A = - 5, B = 3$
D
$A = - 5, B = - 3$

Solution

Let

I = (\frac{4 e^{x} - 25}{2 e^{x} - 5}) d x

\Rightarrow I = \int (\frac{10 e^{x} - 25 - 6 e^{x}}{2 e^{x} - 5}) d x

\Rightarrow I = \int \frac{5 (2 e^{x} - 5) - 6 e^{x}}{2 e^{x} - 5} d x

\Rightarrow I = \int (5 - \frac{6 e^{x}}{2 e^{x} - 5}) d x

\Rightarrow I = 5 x - 3 \log (2 e^{x} - 5) + C

∴ A = 5 a n d B = - 3

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

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