TS EAMCET · Maths · Definite Integration
If \(\mathrm{I}_n=\int \frac{1}{\left(x^2+1\right)^n} d x\), then \(2 n \mathrm{I}_{n+1}-(2 n-1) \mathrm{I}_n=\)
- A \(\frac{\left(x^2+1\right)^n}{x}+\mathrm{c}\)
- B \(\frac{x}{\left(x^2+1\right)^n}+c\)
- C \(x\left(x^2+1\right)^{n-1}+c\)
- D \(\frac{x}{\left(x^2+1\right)^{n-1}}+\mathrm{c}\)
Answer & Solution
Correct Answer
(B) \(\frac{x}{\left(x^2+1\right)^n}+c\)
Step-by-step Solution
Detailed explanation
\(I_n = \int 1 \cdot (x^2+1)^{-n} dx\) \(I_n = x(x^2+1)^{-n} - \int x(-n(x^2+1)^{-n-1} \cdot 2x) dx\) \(I_n = \frac{x}{(x^2+1)^n} + 2n \int \frac{x^2}{(x^2+1)^{n+1}} dx\) \(I_n = \frac{x}{(x^2+1)^n} + 2n \int \frac{(x^2+1)-1}{(x^2+1)^{n+1}} dx\)…
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