JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} d x=\)
\(\frac{1}{m}\left(\left(\sqrt{1+x^2}+x\right)^n\left(n \sqrt{1+x^2}-x\right)\right)+C\)
where C is the constant of integration and \(m, n \in N\), then \(\mathrm{m}+\mathrm{n}\) is equal to
- A 380
- B 381
- C 379
- D 378
Answer & Solution
Correct Answer
(C) 379
Step-by-step Solution
Detailed explanation
rationalise \(\Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} \times \frac{\left(\sqrt{1+x^2}+x\right)^9}{\left(\sqrt{1+x^2}+x\right)^9} d x \) \( \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{1} d x\) Put…
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