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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 5 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 5  Maths Textbook Solution.

Answers (1)

Answer: -\frac{\sqrt{5}}{15} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c

Hint:  Using Formula \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c

Given:\int \frac{x-1}{3 x^{2}-4 x+3} d x

Solution:

            \int \frac{x-1}{3 x^{2}-4 x+3} d x

  \frac{1}{9} \int \frac{x-1}{x^{2}-\frac{4 x}{3}+1} d x

x-1=A+B \frac{d}{d x}\left(x^{2}-\frac{4 x}{3}+1\right)

x-1=A+B\left(2 x-\frac{4}{3}\right)                                                                                    \left[\begin{array}{l} x=2 B x \Rightarrow B=\frac{1}{2} \\ -1=A-\frac{4}{3} B \Rightarrow-1+\frac{4}{3} \times \frac{1}{2}=A \\ -1+\frac{2}{3}=A \Rightarrow A=\frac{-1}{3} \end{array}\right]

I=\frac{-1}{3 \times 3} \int \frac{d x}{x^{2}-\frac{4}{3} x+1}+\frac{1}{3 \times 2} \int \frac{6 x-4}{x^{2}-\frac{4}{3} x+1} d x

I=\frac{-1}{9} \int \frac{d x}{x^{2}-\frac{4}{3} x+1}+\frac{1}{6} \int \frac{d t}{t}                                                                        \left[\begin{array}{c} \text { Let, } x^{2}-\frac{4}{3} x+1=t \\ \left(2 x-\frac{4}{3}\right) d x=d t \\ \frac{(6 x-4) d x}{3}=d t \end{array}\right]

I=\frac{-1}{9} \int \frac{d x}{x^{2}-\frac{4}{3} x+\frac{4}{9}+1-\frac{4}{9}}+\frac{1}{2} \int \frac{d t}{t}

I=-\frac{1}{9} \int \frac{d x}{\left(x-\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{5}}{8}\right)^{2}}+\frac{1}{2} \int \frac{d t}{t}                                        \left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c\right]

I=\frac{-1}{9} \times \frac{1}{\frac{\sqrt{5}}{3}} \tan ^{-1}\left|\frac{x-\frac{2}{3}}{\frac{\sqrt{5}}{3}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c

I=\frac{-1}{3 \sqrt{5}} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c

I=\frac{-3 \sqrt{5}}{3 \sqrt{5} \times 3 \sqrt{5}} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c

I=\frac{-\sqrt{5}}{15} \tan ^{-1}\left|\frac{3 x-2}{\sqrt{5}}\right|+\frac{1}{2} \log \left|x^{2}-\frac{4}{3} x+1\right|+c

Posted by

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