JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of the integral \(\int \limits_1^2\left(\frac{t^4+1}{t^6+1}\right) d t\) is \(..........\).
- A \(\tan ^{-1} \frac{1}{2}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
- B \(\tan ^{-1} 2-\frac{1}{3} \tan ^{-1} 8+\frac{\pi}{3}\)
- C \(\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
- D \(\tan ^{-1} \frac{1}{2}-\frac{1}{3} \tan ^{-1} 8+\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(I=\int \limits_1^2\left(\frac{t^4+1}{t^6+1}\right) d t\) \(=\int \limits_1^2 \frac{\left(t^4+1-t^2\right)+t^2}{\left(t^2+1\right)\left(t^4-t^2+1\right)} d t\) \(=\int \limits_1^2\left(\frac{1}{t^2+1}+\frac{t^2}{t^6+1}\right) d t\)…
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