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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 10 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 10 Maths Textbook Solution.

Answers (2)

Answer: \sqrt{x^{2}+x+1}-\frac{1}{2} \log \left|\frac{2 x+1}{2}+\sqrt{x^{2}+x+1}\right|+c

Given: \int \frac{x}{\sqrt{x^{2}+x+1}} d x

Hint: Simplify the given (f(x))^{n}

Solution:

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

          \begin{aligned} &I=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}} d x \\ &I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+x+1}}{\frac{1}{2}}\right]-\frac{1}{2} \log \left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|+c \end{aligned}

         \left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{array}\right]

         I=\sqrt{x^{2}+x+1}-\frac{1}{2} \log \left|\frac{2 x+1}{2}+\sqrt{x^{2}+x+1}\right|+c

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Answer:\sqrt{x^{2}+x+1}-\frac{1}{2} \log \left|\frac{2 x+1}{2}+\sqrt{x^{2}+x+1}\right|+c

Given:\int \frac{x}{\sqrt{x^{2}+x+1}} d x

Hint: Simplify the given (f(x))^{n}

Solution:

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

         I=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}} d x

       I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+x+1}}{\frac{1}{2}}\right]-\frac{1}{2} \log \left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|+c

     \begin{aligned} &{\left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{array}\right]} \\ &I=\sqrt{x^{2}+x+1}-\frac{1}{2} \log \left|\frac{2 x+1}{2}+\sqrt{x^{2}+x+1}\right|+c \end{aligned}

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