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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 9 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 9 Maths Textbook Solution.

Answers (1)

Answer:

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

Given:

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

Hint:

The given integral function can be converted into

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

Solution

\begin{aligned} I &=\int \frac{(x-1)^{2}}{x^{4}+x^{2}+1} d x \\ &=\int \frac{x^{2}-2 x+1}{x^{4}+x^{2}+1} d x \\ &=\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x+\int \frac{-2 x}{x^{4}+x^{2}+1} d x \end{aligned}

\begin{aligned} \text { let } x^{2} &=t \\ 2 x d x &=d t \\ I &=\int \frac{x^{2}\left(1+\frac{1}{x^{2}}\right)}{x^{2}\left(x^{2}+1+\frac{1}{x^{2}}\right)} d x+\int \frac{-d t}{t^{2}+t+1} \text { (substituting } t \text { as }x^{2}) \\ &=\int \frac{\left(1+\frac{1}{x^{2}}\right)}{\left(x-\frac{1}{x^{2}}\right)^{2}+2+1} d x-\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}+1} d t \end{aligned}

\begin{aligned} &\text { let } x-\frac{1}{x}=p \\ &\left(1+\frac{1}{x^{2}}\right) d x=d p \\ &I=\int \frac{d p}{p^{2}+3}-\int \frac{d t}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}\left(\text { substituting } p \text { as } x-\frac{1}{x}\right) \\ &=\int \frac{1}{p^{2}+(\sqrt{3})^{2}} d p-\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d t \end{aligned}

Using the identity

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

\begin{aligned} I &=\frac{1}{\sqrt{3}} \tan ^{-1} \frac{p}{\sqrt{3}}-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \\ &\left.=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right) \text { (substituting } p \text { as }\left(x-\frac{1}{x}\right) \text { and } t \text { as } x^{2}\right) \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)-\frac{2}{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c \end{aligned}

So, the required value of the given integral is,

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

where c is integrating constant.

 

Posted by

infoexpert27

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