CUET · MATHS · PYQ PAPER 2023
The value of \(\int \frac{1}{\sqrt{x^2+8 x+9}} d x\) is(Given \(C\) constant of integration):
- A \(\log \left[(x+4)+\sqrt{x^2+8 x+9}\right]+C\)
- B \(\log \left[(x+4)-\sqrt{x^2+8 x+9}\right]+C\)
- C \(\log \left[(x-4)+\sqrt{x^2-8 x+9}\right]+C\)
- D \(\log \left[(x-4)-\sqrt{x^2+8 x-9}\right]+C\)
Answer & Solution
Correct Answer
(A) \(\log \left[(x+4)+\sqrt{x^2+8 x+9}\right]+C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{1}{\sqrt{x^2+8 x+9}} d x = \int \frac{1}{\sqrt{(x+4)^2 - 7}} d x\) \(= \log \left|(x+4)+\sqrt{(x+4)^2 - 7}\right|+C\) \(= \log \left[(x+4)+\sqrt{x^2+8 x+9}\right]+C\)
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