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JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int {\frac{{xdx}}{{2 - {x^2} + \sqrt {2 - {x^2}} }}} \) equals
- A \( \log \left| {1 + \sqrt {2 + {x^2}} } \right| + c\)
- B \( - \log \left| {1 + \sqrt {2 - {x^2}} } \right| + c\)
- C \( - x\log \left| {1 - \sqrt {2 - {x^2}} } \right| + c\)
- D \( x\log \left| {1 - \sqrt {2 + {x^2}} } \right| + c\)
Answer & Solution
Correct Answer
(B) \( - \log \left| {1 + \sqrt {2 - {x^2}} } \right| + c\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\int \frac{x d x}{2-x^{2}+\sqrt{2-x^{2}}}\) Put \(t=\sqrt{2-x^{2}}, \frac{d t}{d x}\) \(=\frac{1}{2 \sqrt{2-x^{2}}} \cdot(-2 x)\) \(\Rightarrow-t d t=x d x\) \(\therefore \) \(I=\int \frac{(-t) d t}{t^{2}+t}\) \(=-\int \frac{1}{t+1} d t\) \(=-\log |t+1|\)…
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