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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 18 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 18 maths Textbook Solution.

Answers (2)

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

Hint : Let  x^{2}=t

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

Solution: \int \frac{x\left(x^{2}-3\right)}{\left(x^{2}\right)^{2}+2 x^{2}-4} d x

                Let  x^{2}=t

                   x d x=\frac{d t}{2}

         \frac{1}{2} \int \frac{(t-3)}{t^{2}+2 t-4} d t

        Multiply and divide by 2

          \begin{aligned} &=\frac{1}{4} \int \frac{2 t-6}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2-8}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-\frac{1}{4} \int \frac{8}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{t^{2}+2 t+1-5} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{(t+1)^{2}-\sqrt{5}^{2}} d t \end{aligned}

I=\frac{1}{4}\log \left | t^{2}+2t-4 \right |-2\times \frac{1}{2\sqrt{5}}\log (\frac{t+1-\sqrt{5}}{t+1+\sqrt{5}})+c

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

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

Hint : Let  x^{2}=t

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

Solution: \int \frac{x\left(x^{2}-3\right)}{\left(x^{2}\right)^{2}+2 x^{2}-4} d x

                Let  x^{2}=t

                   x d x=\frac{d t}{2}

         \frac{1}{2} \int \frac{(t-3)}{t^{2}+2 t-4} d t

        Multiply and divide by 2

          \begin{aligned} &=\frac{1}{4} \int \frac{2 t-6}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2-8}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-\frac{1}{4} \int \frac{8}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{t^{2}+2 t-4} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{t^{2}+2 t+1-5} d t \\ &=\frac{1}{4} \int \frac{2 t+2}{t^{2}+2 t-4} d t-2 \int \frac{1}{(t+1)^{2}-\sqrt{5}^{2}} d t \end{aligned}

  First integration and differentiate both sides
\text { let, } \quad \begin{aligned} &t^{2}+2 t-4=u \\ &(2 t+2) d t=d u \end{aligned}

I=\frac{1}{4} \int \frac{d u}{u}-2 \int \frac{1}{(t+1)^{2}-\sqrt{5}^{2}} d t                                                        \left[\int \frac{1}{x} d x=\log |x| \& \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\right]

I=\frac{1}{4} \log |u|-2 \frac{1}{2 \sqrt{5}} \log \left|\frac{t+1-\sqrt{5}}{t+1+\sqrt{5}}\right|+c

                    Where c is integration constant

                    Put u=t^{2}+2 t-4 \& t=x^{2}

                                       \begin{aligned} &I=\frac{1}{4} \log \left|t^{2}+2 t-4\right|-\frac{1}{\sqrt{5}} \log \left|\frac{t+1-\sqrt{5}}{t+1+\sqrt{5}}\right|+c \\ &I=\frac{1}{4} \log \left|x^{4}+2 x^{2}-4\right|-\frac{1}{\sqrt{5}} \log \left|\frac{x^{2}+1-\sqrt{5}}{x^{2}+1+\sqrt{5}}\right|+c \end{aligned}

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