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  • Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 7

Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 7

Answers (1)

Answer:

        \frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}+1}{\sqrt{2}}\right)+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

        \int \frac{x}{x^{4}+2x^{2}+3}dx

Solution:

Let\: \: \: I=\int \frac{x}{x^{4}+2x^{2}+3}dx

Put\: \: x^{2}=t\Rightarrow 2x\, dx=dt\Rightarrow x\: dx=\frac{dt}{2}

Then\: \: \: I=\int \frac{1}{t^{2}+2t+3}\frac{dt}{2}

                    \begin{aligned} &=\frac{1}{2} \int \frac{1}{t^{2}+2 t+3} d t \\ &=\frac{1}{2} \int \frac{1}{t^{2}+2 \cdot t \cdot 1+1^{2}-1^{2}+3} d t \\ &=\frac{1}{2} \int \frac{1}{(t+1)^{2}-1+3} d t \\ &=\frac{1}{2} \int \frac{1}{(t+1)^{2}+2} d t \\ &=\frac{1}{2} \int \frac{1}{(t+1)^{2}+(\sqrt{2})^{2}} d t \end{aligned}

Put\: \: u=t+1\Rightarrow du=dt

Then\: \: \begin{aligned} I=\frac{1}{2} \int \frac{1}{u^{2}+(\sqrt{2})^{2}} d u \end{aligned}

                    \begin{array}{ll} =\frac{1}{2} \times \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+C & {\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]} \\ \\=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+C \quad & {[\because u=t+1]} \\ \\=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}+1}{\sqrt{2}}\right)+C & {\left[\because t=x^{2}\right]} \end{array}

 

 

 

 

Posted by

Gurleen Kaur

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