JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\int\limits_0^{\frac{1}{2}} {\frac{{\ln \,\left( {1 + 2x} \right)}}{{1 + 4{x^2}}}} dx\) , equals
- A \(\frac{\pi }{4}\,\ln \,2\)
- B \(\frac{\pi }{8}\,\ln \,2\)
- C \(\frac{\pi }{16}\,\ln \,2\)
- D \(\frac{\pi }{32}\,\ln \,2\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi }{16}\,\ln \,2\)
Step-by-step Solution
Detailed explanation
Let \(I = \int\limits_0^{1/2} {\frac{{\ln (1 + 2x)}}{{\ln (1 + 2x)}}} dx\) or \(\int\limits_0^{1/2} {\frac{{\ln (1 + 2x)}}{{1 + {{(2x)}^2}}}} dx\) Put \(2 x=\tan \theta\) \(\therefore \frac{2 d x}{d \theta}=\sec ^{2} \theta\) or \(d x=\frac{\sec ^{2} \theta d \theta}{2}\) also…
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