The value of the integral ∫-log ₑ 2^(log ₑ 2) eˣ(log ₑ(eˣ+√(1+e²ˣ))) d x is equal to :

Solution

Maths sir · 4 steps

Put t = eˣ first. The log then becomes sinh⁻¹ t, and parts finishes it.

1. Substitute. Let t = eˣ, so dt = eˣ dx. The limits x = -ln 2 and x = ln 2 become t = 1/2 and t = 2. The integral is now ∫_(1/2)² ln(t+√(1+t²)) dt.
2. Integrate by parts. Take ln(t+√(1+t²)) as the first function and 1 as the second. Its derivative is 1/√(1+t²), and ∫ t/√(1+t²) dt = √(1+t²).

   ∫ ln(t+√(1+t²)) dt = t ln(t+√(1+t²)) - √(1+t²)

3. Put in limits. At t = 2 this is 2 ln(2+√5) - √5. At t = 1/2 it is 1/2 ln(1+√5)/2 - √5/2. Subtracting gives 2 ln(2+√5) - 1/2 ln(1+√5) + 1/2 ln 2 - √5/2.
4. So. Combining the logs gives ln((√2 (2+√5)²)/√(1+√5)) - √5/2, which is option (b).

The idea to keep. When eˣ sits next to a function of eˣ, substitute t = eˣ straight away. Also remember that ln(t+√(1+t²)) integrates by parts to t ln(…) - √(1+t²).

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