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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 5 maths textbook solution

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        \frac{1}{36} \log \left|\frac{2 e^{3 x}-3}{2 e^{3 x}+3}\right|+C


Use substitution method as well as special integration formula to solve this type of problem


        \int \frac{e^{3 x}}{4 e^{6 x}-9} d x


Let\: \: \int \frac{e^{3 x}}{4 e^{6 x}-9} d x

Put \: \: \: e^{3x}=t\Rightarrow 3e^{3x}dx=dt\Rightarrow e^{3x}dx=\frac{dt}{3}

Then\: \: \: I=\int \frac{1}{4 t^{2}-9} \frac{d t}{3}=\int \frac{1}{4\left(t^{2}-\frac{9}{4}\right)} \frac{d t}{3}

                    =\frac{1}{4 \times 3} \int \frac{1}{t^{2}-\frac{9}{4}} d t=\frac{1}{4 \times 3} \int \frac{1}{t^{2}-\left(\frac{3}{2}\right)^{2}} d t

                    =\frac{1}{4 \times 3}\left[\frac{1}{2 \times \frac{3}{2}} \log \left|\frac{t-\frac{3}{2}}{t+\frac{3}{2}}\right|\right]+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]

                    \begin{aligned} &=\frac{1}{4 \times 3 \times 3} \log \left|\frac{\frac{2 t-3}{2}}{\frac{2 t+3}{2}}\right|+C \\ \\ &=\frac{1}{36} \log \left|\frac{2 t-3}{2 t+3}\right|+C \\\\ &=\frac{1}{36} \log \left|\frac{2 e^{3 x}-3}{2 e^{3 x}+3}\right|+C \quad\left[\because t=e^{3 x}\right] \end{aligned}

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Gurleen Kaur

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