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

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

Answers (1)

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}

 

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