The value of the integral ∫-1² log ₑ(x+√(x²+1)) d x is

Solution

Maths sir · 4 steps

The integrand is an odd function, so the part from -1 to 1 cancels and only 1 to 2 is left.

1. Spot the odd function. Let f(x) = ln(x+√(x²+1)). Then f(-x) = ln(√(x²+1)-x) = ln 1/(x+√(x²+1)) = -f(x), so f is odd. Hence ∫₋₁¹ f(x) dx = 0, and the integral equals ∫₁² f(x) dx.
2. Integrate by parts. Take f(x) as the first function and 1 as the second. Since f'(x) = 1/√(x²+1), the remaining integral is ∫ x/√(x²+1) dx = √(x²+1).

   ∫ ln(x+√(x²+1))dx = x ln(x+√(x²+1)) - √(x²+1) + C

3. Apply the limits. At x = 2 it gives 2 ln(2+√5) - √5. At x = 1 it gives ln(1+√2) - √2. Subtracting gives √2 - √5 + 2 ln(2+√5) - ln(1+√2).
4. So. 2 ln(2+√5) = ln(2+√5)² = ln(9+4√5), so the value is √2-√5+ln((9+4√5)/(1+√2)). That is option (d).

The idea to keep. Before integrating over a range like -a to b, check whether the function is odd. The symmetric part -a to a vanishes, and you only work on what is left.

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