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  • Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.15 Question 4

Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.15 Question 4

Answers (1)

Answer:

            \frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C

Hint:

            To solve this problem use special integration formula

Given:

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

Solution:

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

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

            =\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{1}{16}-\frac{1}{2}} d x

            \begin{aligned} &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\left(\frac{1+8}{16}\right)} d x \\ & \end{aligned}

            =\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}} d x

Put   x-\frac{1}{4}=t \Rightarrow d x=d t

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

            =\frac{1}{2} \cdot \frac{1}{2 \times \frac{3}{4}} \log \left|\frac{t-\frac{3}{4}}{t+\frac{3}{4}}\right|+C                                 \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}{3} \log \left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+C \text { }                                                 \quad\left[\because t=x-\frac{1}{4}\right]

            \begin{aligned} &=\frac{1}{3} \log \left|\frac{\frac{4 x-1-3}{4}}{\frac{4 x-1+3}{4}}\right|+C \\ & \end{aligned}

            =\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C

Posted by

infoexpert27

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