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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 8 Maths Textbook Solution.

Answers (1)

Answer:

\frac{x^{3}}{3}-\tan ^{-1} x+c

Given:

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

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

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

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

=\frac{x^{3}}{3}-\tan ^{-1} x+c

 

 

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