Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18 .19 Question 17 maths Textbook Solution.

Answer: $\frac{1}{4} \ln \left|x^{4}+x^{2}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c$

Hint: You know about the integration of $\int x^{3}$

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

Solution:  $\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$

\begin{aligned} &\int \frac{x^{3}}{x^{4}+x^{2}+1} d x \\ &=\frac{1}{2} \int \frac{2 x^{3}}{x^{4}+x^{2}+1} d x \\ &=\frac{1}{2} \int \frac{x^{2}}{x^{4}+x^{2}+1} 2 x d x \end{aligned}

Let $x^{2}=t$

$2 x d x=d t$

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

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

Let, \begin{aligned} t^{2}+2 t+1 &=u \\ 2 t+1 d t &=d u \end{aligned}

\begin{aligned} &=\frac{1}{4} \int \frac{d u}{u}-\int \frac{d t}{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(t+\frac{1}{2}\right)^{2}} \\ &=\frac{1}{4}\left[\ln u-\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]+c \end{aligned}

\begin{aligned} &=\frac{1}{4}\left(\ln \left(t^{2}+t+1\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)\right)+c \\ &=\frac{1}{4} \ln \left|x^{4}+x^{2}+1\right|-\frac{1}{4} \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c \\ &=\frac{1}{4} \ln \left|x^{4}+x^{2}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c \end{aligned}