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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 3 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 3 Maths Textbook Solution.

Answers (1)

Answer: The required value of the integral is,

I=\frac{1}{3 \sqrt{2}}\left(\frac{x-\frac{9}{x}}{3 \sqrt{2}}\right)+c

Hint: Use the formula\int \frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+c

Given:\int \frac{\left(x^{2}+9\right)}{\left(x^{4}+81\right)} d x

Solution: The equation can be written as

I=\int \frac{1+\frac{9}{x^{2}}}{x^{2}+\frac{81}{x^{2}}} d x

=\int \frac{1+\frac{9}{x^{2}}}{\left(x-\frac{9}{x}\right)^{2}+18} d x

 Let, x-\frac{9}{x}=t then,

Differentiating the above equation both sides.

\left(1+\frac{9}{x^{2}}\right) d x=d t then the obtained equation become

\int \frac{d t}{t^{2}+(3 \sqrt{2})^{2}}

 Now, using the identity rule,

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

Now, re-substituting t=x-\frac{9}{x} then,

\begin{aligned} I &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{9}{x}}{3 \sqrt{2}}\right)+c \\ &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-9}{3 \sqrt{2} x}\right)+c \end{aligned}

Where c is the integrating constant.

Posted by

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