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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 4 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 4 Maths Textbook Solution.

Answers (1)

Answer:  The required value of the integral is,

I=\frac{1}{2}\left[\frac{1}{\sqrt{3}}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{1}{2} \log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right]

Hint:  Use the identity  formula

\int \frac{1}{x^{2}+1} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \text { and } \int \frac{d x}{x^{2}-1}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c

Given:

\int \frac{1}{x^{4}+x^{2}+1} d x

Solution: The given equation can be written as,

I=\int \frac{\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x                                [dividing x^{2} in both denminator and numerator]

\begin{aligned} &=\frac{1}{2} \int \frac{1+\frac{1}{x^{2}}+\frac{1}{x^{2}}-1}{x^{2}+1+\frac{1}{x^{2}}} d x \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x+\int \frac{-1+\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1} d x\right] \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d x+\int \frac{-1+\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1} d x\right] \end{aligned}                                [Making the perfect square as\left ( a+b \right )^{2}  ]

 Re-writing the given equation as,

\begin{aligned} &\text { Let, } x-\frac{1}{x}=t \text { and } x+\frac{1}{x}=z \text { so } \\ &\begin{aligned} \left(1+\frac{1}{x^{2}}\right) d x &=d t \text { and }\left(1-\frac{1}{x^{2}}\right) d x=d z \\ I &=\frac{1}{2}\left[\int \frac{d t}{t^{2}+3}-\int \frac{d z}{z^{2}-1}\right] \end{aligned} \end{aligned}

On using identity,

\begin{aligned} \int \frac{1}{x^{2}+1} d x &=\tan ^{-1} x \text { and } \int \frac{d z}{z^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c \\ I &=\frac{1}{2}\left[\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)-\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|\right] \end{aligned}

Now, re-substituting t=x-\frac{1}{x}and\: z=x+\frac{1}{x}

The required value of the integral is,

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

Where c is the integrating constant.

Posted by

infoexpert27

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