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  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 33 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 33 maths.

Answers (1)

Answer: (\log \sqrt{3} )

Hint: We use indefinite integral formula and the limits to solve this integral.

Given: \int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x

Solution: \int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x=-\int_{0}^{1} \frac{x^{2}-1}{x^{4}+x^{2}+1} d x

Dividing the num. and denom. by x2  than

\begin{aligned} &\int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x=-\int_{0}^{1} \frac{x^{2}-1}{x^{2}+x^{2}+1}{x^{2}} d x \mid \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+1+1-1} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+2-1} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+2 x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}-1} d x \end{aligned}

\begin{aligned} &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+2 x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}-1} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-(1)^{2}} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1^{2}} d x \end{aligned}

put x+\frac{1}x{}=t   then, (1-\frac{1}{x^2}) dx=dt

\begin{aligned} &\therefore \int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1^{2}} d x \\ &=-\int_{0}^{1} \frac{1}{t^{2}-1} d t \\ &=-\frac{1}{2 \times 1}\left[\log \left|\frac{t-1}{t+1}\right|\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int \frac{1}{x^{2}-a^{2}} \mathrm{dx}=\frac{1}{2} \log \left|\frac{x-a}{x+a}\right|+c\right] \end{aligned}

\begin{aligned} &=-\frac{1}{2}\left[\log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right]_{0}^{1} \quad\left[\because t=x+\frac{1}{x}\right] \\ &=-\frac{1}{2}\left[\log \left|\frac{\frac{x^{2}+1-x}{x}}{\frac{x^{2}+1+x}{x}}\right|\right]_{0}^{1} \\ &=-\frac{1}{2}\left[\log \left|\frac{x^{2}-x+1}{x^{2}+x+1}\right|\right]_{0}^{1} \\ &=-\frac{1}{2}\left[\log \left|\frac{1^{2}-1+1}{1^{2}+1+1}\right|-\log \left|\frac{0^{2}-0+1}{0^{2}+0+1}\right|\right] \\ &=-\frac{1}{2}\left[\log _{3}^{1}-\log \frac{1}{1}\right] \end{aligned}

\begin{aligned} &=-\frac{1}{2}\left[\log _{3}^{1}-\log 1\right] \\ &=-\frac{1}{2}\left[\log _{3}^{1}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \log 1=0] \\ &=-\frac{1}{2} \cdot \log _{3}^{\frac{1}{3}} \\ &=-\frac{1}{2} \cdot \log 3^{-1} \\ &=\log 3^{-1\left(\frac{-1}{2}\right)} \end{aligned}

=\log 3^{\frac{1}{2}}

=\log\sqrt{3}

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