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  • Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.15 Question 3

Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.15 Question 3

Answers (1)

Answer:

            \frac{-1}{\sqrt{5}} \log \left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C

Hint:

            To solve this problem use special integration formula

Given:

            \int \frac{1}{1+x-x^{2}} d x

Solution:

Let    I=\int \frac{1}{1+x-x^{2}} d x=-\int \frac{1}{x^{2}-x-1} d x

            \begin{aligned} &=-\int \frac{1}{x^{2}-2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-1} d x \\ & \end{aligned}

            =-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-1} d x

            \begin{aligned} &=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1+4}{4}\right)} d x \\ & \end{aligned}

            =-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}} d x

Put    x-\frac{1}{2}=t \Rightarrow d x=d t

Then    I=-\int \frac{1}{t^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d t

            =\frac{-1}{2 \cdot \frac{\sqrt{5}}{2}} \log \left|\frac{t-\frac{\sqrt{5}}{2}}{t+\frac{\sqrt{5}}{2}}\right|+C \text { }                                 \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]

            =\frac{-1}{\sqrt{5}} \log \left|\frac{x-\frac{1}{2}-\frac{\sqrt{5}}{2}}{x-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|+C                                           \quad\left[\because t=x-\frac{1}{2}\right]

            \begin{aligned} &=\frac{-1}{\sqrt{5}} \log \left|\frac{\frac{2 x-1-\sqrt{5}}{2}}{\frac{2 x-1+\sqrt{5}}{2}}\right|+C \\ & \end{aligned}

            =\frac{-1}{\sqrt{5}} \log \left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C

 

Posted by

infoexpert27

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