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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.15 Question 1 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.15 Question 1 Maths Textbook Solution.

Answers (1)

Answer:

            \frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C

Hint:

            To solve this problem use special integration formula

Given:

            \int \frac{1}{4 x^{2}+12 x+5} d x

Solution:

Let       I=\int \frac{1}{4 x^{2}+12 x+5} d x=\frac{1}{4} \int \frac{1}{x^{2}+\frac{12 x}{4}+\frac{5}{4}} d x

\begin{aligned} &=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{4}} d x=\frac{1}{4} \int \frac{d x}{x^{2}+2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}} \\\\ &=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{5}{4}} d x=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9-5}{4}\right)} d x \end{aligned}
 

Put     x+\frac{3}{2}=t \Rightarrow d x=d t


Then      I=\frac{1}{4} \int \frac{1}{t^{2}-1^{2}} d t

            =\frac{1}{4} \cdot \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\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]

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

            \begin{aligned} &=\frac{1}{8} \log \left|\frac{\frac{2 x+3-2}{2}}{\frac{2 x+3+2}{2}}\right|+C \\\\ &\end{aligned}

             =\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C

 

Posted by

infoexpert27

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