JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .\) Then
- A \(50\,I _{6}-9\,I _{5}= xI _{5}^{\prime}\)
- B \(50\,I _{6}-11\,I _{5}= xI _{5}^{\prime}\)
- C \(50\,I _{6}-9\,I _{5}= I _{5}^{\prime}\)
- D \(50\,I _{6}-11\,I _{5}= I _{5}^{\prime}\)
Answer & Solution
Correct Answer
(A) \(50\,I _{6}-9\,I _{5}= xI _{5}^{\prime}\)
Step-by-step Solution
Detailed explanation
\(I_{n}(x)=\int_{0}^{ x } \frac{ dt }{\left( t ^{2}+5\right)^{ n }}\) Applying integral by parts \(I_{n}(x)=\left[\frac{t}{\left(t^{2}+5\right)^{ n }}\right]_{0}^{ x }-\int_{0}^{ x } n \left( t ^{2}+5\right)^{- n -1} \cdot 2 t ^{2}\)…
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