Question

$\int \frac{\left(4e^x+6e^{-x}\right)}{\left(9e^x-4e^{-x}\right)}dx=$

• A. $\frac{3}{2}x+\left(\frac{35}{35}\right)\log \left|9e^{2x}-4\right|+C$
• B. $\frac{3}{2}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C$
• C. $-\frac{3}{2}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C$
• D. $-\frac{5}{2}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C$
• E. $\frac{5}{2}x+\left(\frac{35}{66}\right)\log \left|9e^{2x}-4\right|+C$

Answer 1

The correct option is C

$-\frac{3}{2}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C$

Explanation for the correct option:

Step1: Expanding the given integral

$$\begin{gathered} \int \frac{\left(4e^x+6e^{-x}\right)}{\left(9e^x-4e^{-x}\right)}dx=\int \frac{\left(4e^x+{\displaystyle \frac{6}{e^x}}\right)}{\left(9e^x-{\displaystyle \frac{4}{e^x}}\right)}dx \\ =\int \frac{\left({\displaystyle \frac{4e^{2x}+6}{\cancel{e^x}}}\right)}{\left({\displaystyle \frac{9e^{2x}-4}{\cancel{e^x}}}\right)}dx \\ =\int \frac{4e^{2x}+6}{9e^{2x}-4}dx \\ =\frac{4}{9}\int \frac{9e^{2x}+{\displaystyle \frac{6}{4}}\times 9}{9e^{2x}-4}dx\left[\text{take 4 as common and then × and÷ by 9 in numerator}\right] \\ =\frac{4}{9}\int \frac{\left(9e^{2x}-4\right)+4+{\displaystyle \frac{27}{2}}}{9e^{2x}-4}dx\left[\text{Add and subtract by 4 in numerator}\right] \end{gathered}$$$$\begin{gathered} =\frac{4}{9}\left[\int \frac{\left(9e^{2x}-4\right)}{9e^{2x}-4}dx+\left(4+\frac{27}{2}\right)\int \frac{1}{9e^{2x}-4}dx\right] \\ =\frac{4}{9}\left[\int dx+\left(\frac{35}{2}\right)\int \frac{1}{\left(9e^{2x}-4\right)}dx\right]\to \left(i\right) \end{gathered}$$

$$\begin{gathered} \int \frac{1}{9e^{2x}-4}dx=\int \frac{dz}{z\left({\displaystyle \frac{4+z}{9}}\right)}\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{z}\mathbf{=}\mathbf{9}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{-}\mathbf{4}\mathbf{;}\mathbf{\Rightarrow }{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{=}\frac{\mathbf{4}\mathbf{+}\mathbf{z}}{\mathbf{9}}\mathbf{,}\mathbf{}\mathbf{\Rightarrow }\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{d}\mathbf{x}\mathbf{=}\left(\frac{1}{9}\right)\mathbf{dz}\mathbf{\right]} \\ =\frac{1}{2}\int \frac{dz}{z\left(4+z\right)} \\ =\frac{1}{2}\int \frac{dz}{z\left(4+z\right)} \end{gathered}$$

Step 2: Integrating by partial fraction

$$\begin{gathered} \int \frac{dz}{z\left(4+z\right)}=\int \frac{A}{z}dz+\int \frac{B}{\left(4+z\right)}dz \\ \text{ie,}A\left(4+z\right)+Bz=1 \\ \text{where}z=-4\Rightarrow B=\frac{-1}{4} \\ \text{where}z=0\Rightarrow A=\frac{1}{4} \\ \Rightarrow \frac{1}{2}\int \frac{dz}{z\left(4+z\right)}=\frac{1}{2}\left[\frac{1}{4}\int \frac{1}{z}dz-\frac{1}{4}\int \frac{1}{\left(4+z\right)}dz\right] \\ =\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)\left[\int \frac{1}{z}dz-\int \frac{1}{\left(4+z\right)}dz\right] \end{gathered}$$

Substituting in equation $\left(i\right)$ we get,

$$\begin{gathered} =\frac{4}{9}\left[\int dx+\left(\frac{35}{2}\right)\frac{1}{8}\left[\int \frac{1}{z}dz-\int \frac{1}{\left(4+z\right)}dz\right]\right] \\ =\frac{4}{9}\int dx+\left(\frac{35}{36}\right)\int \frac{1}{z}dz-\left(\frac{35}{36}\right)\int \frac{1}{\left(4+z\right)}dz \end{gathered}$$

$$\begin{gathered} =\frac{4}{9}x+\left(\frac{35}{36}\right)\log \left|z\right|-\left(\frac{35}{36}\right)\log \left|4+z\right|+C \\ =\frac{4}{9}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|-\left(\frac{35}{36}\right)\log \left|9e^{2x}\right|+C \\ =\frac{4}{9}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|-\left(\frac{35}{36}\right)\log \left|9\right|-\left(\frac{35}{36}\right)\log \left|e^{2x}\right|+C\mathbf{[}\mathbf{\because }\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{b}\mathbf{=}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{b}\mathbf{]} \\ =\frac{4}{9}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|-\left(\frac{35}{36}\right)2x+C\mathbf{[}\mathbf{\because }\mathbf{lne}\mathbf{=}\mathbf{1}\mathbf{]} \\ =\frac{4}{9}x-\left(\frac{35}{18}\right)x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C \\ =\left(-\frac{27}{18}\right)x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C \\ =-\frac{3}{2}x+\left(\frac{35}{36}\right)\log \left|9e^{2x}-4\right|+C \end{gathered}$$

Hence, option (C) is the correct answer.