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

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}