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Question

Mathematics Question on Differential equations

∫dx(x2+1)(x2+4)=∫\frac{dx}{(x2+1) (x2+4)} =

A

13Tan−1x+16Tan−1(x2)+c\frac{1}{3}Tan^{-1}x +\frac{1}{6}Tan^{-1}(\frac{x}{2})+c

B

13Tan−1x−13Tan−1(x2)+c\frac{1}{3}Tan^{-1}x -\frac{1}{3}Tan^{-1}(\frac{x}{2})+c

C

13Tan−1x+13Tan−1(x2)+c\frac{1}{3}Tan^{-1}x +\frac{1}{3}Tan^{-1}(\frac{x}{2})+c

D

13Tan−1x−16Tan−1(x2)+c\frac{1}{3}Tan^{-1}x -\frac{1}{6}Tan^{-1}(\frac{x}{2})+c

Answer

13Tan−1x−16Tan−1(x2)+c\frac{1}{3}Tan^{-1}x -\frac{1}{6}Tan^{-1}(\frac{x}{2})+c

Explanation

Solution

The correct option is (D) 13Tan−1x−16Tan−1(x2)+c\frac{1}{3}Tan^{-1}x -\frac{1}{6}Tan^{-1}(\frac{x}{2})+c