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  • Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 14

Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 14

Answers (1)

Answer:

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

Hint:

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

GIven:

        \int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x

Solution:

Let\: \: I=\int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x

Let\: \: e^{x}=t\Rightarrow e^{x}dx=dt

Then\: \: I=\int \frac{1}{\left(1+t\right)\left(2+t\right)} d t

                \begin{aligned} &=\int \frac{1}{t^{2}+2 t+t+2} d t \\ &=\int \frac{1}{t^{2}+3 t+2} d t \\ &=\int \frac{1}{t^{2}+2 \cdot t \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\frac{9}{4}+2} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\left(\frac{9-8}{4}\right)} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\frac{1}{4}} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \end{aligned}

Put\: \: t+\frac{3}{2}=u\Rightarrow dt=du

Then\: \: I=\int \frac{1}{u^{2}-\left(\frac{1}{2}\right)^{2}} d u

                \begin{array}{ll} =\frac{1}{2 \times \frac{1}{2}} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\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] \\ \\=\log \left|\frac{t+\frac{3}{2}-\frac{1}{2}}{t+\frac{3}{2}+\frac{1}{2}}\right|+C \quad\left[\because t+\frac{3}{2}=u\right] \\ \\=\log \left|\frac{2 t+2}{2 t+4}\right|+C \\ \\=\log \left|\frac{t+1}{t+2}\right|+C \\ \end{array}

                =\log \left|\frac{e^{x}+1}{e^{x}+2}\right|+C\: \: \: \: \: \: \: [\because t=e^{x}]

Posted by

Gurleen Kaur

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