The value of the
integral ∫₋₁¹ log ( x + √(x² + 1) )dx is :

Solution

Maths sir · 4 steps

Limits from -1 to 1? First check if the function is odd. Here it is, so the integral is 0.

1. Check the symmetry. Let f(x) = log(x + √(x²+1)). Then f(-x) = log(√(x²+1) - x).
2. Rationalise. The two arguments multiply to (√(x²+1))² - x² = 1. So √(x²+1) - x = 1/(x + √(x²+1)), and its log is the negative of the original log.

   f(-x) = log 1/(x + √(x²+1)) = -f(x)

3. Use the odd property. For an odd function, ∫₋ₐᵃ f(x) dx = 0, because the area on the left of the origin cancels the area on the right. The argument x + √(x²+1) is always positive, so f is defined on all of [-1, 1].
4. So. The integral equals 0, which is option (b).

The idea to keep. Symmetric limits -a to a are a signal: test f(-x) before integrating anything. log(x + √(x²+1)) is a standard odd function, so remember it.

Answer from the NTA final answer key. Solution by PrepWiser.