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AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 70 of 160

Marks: +1, -0

\int\,\frac{3 e^{x} - 7 e^{-x}}{7 e^{x} + 3 e^{-x}}\,dx = K x + L \log\left(e^{-2x} + \frac{7}{3}\right) + C

K+L=

$\frac{-3}{38}$
$\frac{21}{38}$
$\frac{38}{21}$
$\frac{-38}{3}$

Solution: 👈: Video Solution

70. (c)

I = \int \frac{3 e^x - 7 e^x}{7 e^x + 3 e^{-x}}\,dx

= 3 \int \frac{e^x - \frac{7}{3} e^{-x}}{7 e^x + 3 e^{-x}}\,dx

= \frac{3}{7} \int \frac{7 e^x - \frac{49}{3} e^{-x}}{7 e^x + 3 e^{-x}}\,dx

= \frac{3}{7} \int \frac{(7 e^x+3 e^{-x}) - \frac{58}{3} e^{-x}}{7 e^x+3 e^{-x}}\,dx

= \frac{3}{7} \int dx - \frac{3}{7} \times \frac{58}{3} \int \frac{e^{-x}}{7 e^x+3 e^{-x}}\,dx

I = \frac{3x}{7} - \frac{58}{7} \int \frac{e^{-2x}}{7+3 e^{-2x}}\,dx

Put

t = 7 + 3 e^{-2x}

\Rightarrow dt = -6 e^{-2x} dx

I = \frac{3x}{7} - \frac{58}{7} \left(-\frac{1}{6}\right) \int \frac{dt}{t}

I = \frac{3x}{7} + \frac{29}{21} \ln |t| + C

I = \frac{3x}{7} + \frac{29}{21} \ln (7+3 e^{-2x}) + C .

\Rightarrow K = \frac{3}{7}

and

L = \frac{29}{21}

\Rightarrow K+L = \frac{3}{7} + \frac{29}{21} = \frac{9+29}{21} = \frac{38}{21}

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