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  • Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 12

Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 12

Answers (1)

Answer:-

The answer is
I=\frac{x^{2}}{4} \sqrt{x^{4}+1}+\frac{1}{4} \log \left|x+x^{2} \sqrt{x^{4}+1}\right|+c

Hints:-

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}

 

Given:-

I=\int x \sqrt{x^{4}+1} d x

Solution:-

Let,
I=\int x \sqrt{x^{4}+1} d x

Let, x^{2}= t

Differentiating both sides,

\begin{aligned} &\Rightarrow 2 x d x=d t \\\\ &\Rightarrow x d x=\frac{1}{2} d t \end{aligned}

 

Substituting x^{2} with t, we have

\begin{aligned} &I=\frac{1}{2} \int \sqrt{t^{2}+1} d t \\\\ &I=\frac{1}{2} \int \sqrt{t^{2}+1^{2}} d t \end{aligned}

 

As I match with the form

\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &I=\frac{1}{2}\left\{\frac{t}{2} \sqrt{t^{2}+1}+\frac{1}{2} \log \left|t+\sqrt{t^{2}+1}\right|\right\}+c \\\\ &I=\frac{t}{4} \sqrt{t^{2}+1}+\frac{1}{4} \log \left|t+\sqrt{t^{2}+1}\right|+c \end{aligned}

Putting the value of t back.

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

Posted by

infoexpert27

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