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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 46 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 46 Maths Textbook Solution.

Answers (1)

Answer:

I=\frac{1}{\sqrt{17}} \log \left|\frac{8 x+1+\sqrt{17}}{-8 x-1+\sqrt{17}}\right|+c

Given:

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

Hint:

To solve this statement we have to do whole square method.

Solution: 

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

  I=\int \frac{1}{1-4\left(x+\frac{1}{8}\right)^{2}+\frac{1}{16}} d x

I=\int \frac{1}{\frac{17}{6}-4\left(x+\frac{1}{8}\right)^{2}} d x

I=\frac{1}{4} \int \frac{1}{\left(\frac{\sqrt{17}}{8}\right)^{2}-\left(x+\frac{1}{8}\right)^{2}} d x                            \left[\because \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right]

I=\frac{1}{4} \frac{4}{\sqrt{17}} \log \left|\frac{x+\frac{1}{8}+\frac{\sqrt{17}}{8}}{\frac{\sqrt{17}}{8}-x-\frac{1}{8}}\right|+c

I=\frac{1}{\sqrt{17}} \log \left|\frac{8 x+1+\sqrt{17}}{-8 x-1+\sqrt{17}}\right|+c

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