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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 10 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 10 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{2 \sqrt{5}}\left(\frac{x^{2}-1}{\sqrt{5} x}\right)-\frac{1}{2}\left(\frac{x^{2}+1}{x}\right)+c

Given:

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

Hint:

The given integral function can be converted into

\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c

Solution

\begin{aligned} I &=\int \frac{1}{x^{4}+3 x^{2}+1} d x \\ &=\int \frac{\frac{1}{x^{2}}}{x^{2}+3+\frac{1}{x^{2}}} d x \end{aligned}(Numerator and denominator divided by x^{2})

\begin{aligned} &=\frac{1}{2} \int \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+3+\frac{1}{x^{2}}} d x \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+5} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}+1} d x\right] \end{aligned}

\begin{aligned} \text { Assume } t &=x-\frac{1}{x} \text { and } z=x+\frac{1}{x} \\ d t &=\left(1+\frac{1}{x^{2}}\right) d x \text { and } d z=\left(1-\frac{1}{x^{2}}\right) d x \\ &=\frac{1}{2}\left[\int \frac{d t}{t^{2}+5}-\int \frac{d z}{z^{2}+1}\right] \end{aligned}

Using identity \int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\frac{x}{a}+c

I=\frac{1}{2\sqrt{5}}\tan-\left ( \frac{t}{\sqrt{5}} \right )-\frac{1}{2}\tan^{-1}z

Substituting  t \: as\: x-\frac{1}{x}and\: z\: as\: x+\frac{1}{x}

\begin{aligned} &I=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\frac{1}{2} \tan ^{-1}\left(x+\frac{1}{x}\right)+c \\ &\frac{1}{2 \sqrt{5}}\left(\frac{x^{2}-1}{\sqrt{5} x}\right)-\frac{1}{2}\left(\frac{x^{2}+1}{x}\right)+c \end{aligned}

Where c is integrating constant.

Posted by

infoexpert27

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