TS EAMCET · Maths · Indefinite Integration
Given that \[ \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2} \text { and } \frac{d}{d x}\left(\sin h^{-1} x\right)=\frac{1}{\sqrt{1+x^2}} \text {. } \] Then \(\int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x=\)
- A \(\frac{3}{7} x^7-\frac{2}{5} x^5+\frac{1}{3} x^3-2 x+c\)
- B \(\frac{\frac{3}{7} x^7-\frac{2}{5} x^5+\frac{1}{3} x^3-2 x}{\frac{x^3}{3}+x}+c\)
- C \(\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 \tan ^{-1} x+c\)
- D \(\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 \sin \mathrm{h}^{-1} x+c\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 \tan ^{-1} x+c\)
Step-by-step Solution
Detailed explanation
Given \(\int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x\) Now,…
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