JEE Main 2020 · 6 September · evening · Single correct · +4 / −1
The integral ∫₁² eˣ.xˣ( 2 + log ₑx ) dx equals :
- (a) e(4e + 1)
- (b) e(2e – 1)
- (c) e(4e – 1)
- (d) 4e² – 1
Answer and solution
Answer (c) e(4e – 1)
Solution
Maths sir · 4 steps
Spot that (2 + ln x) splits into 1 + (1 + ln x), and the integrand becomes eˣ times f + f'.
- Derivative of x^x. Write xˣ = eˣˡⁿˣ. Then d/dx(xˣ) = xˣ(1 + ln x).
- Split the bracket. Write 2 + ln x = 1 + (1 + ln x). The integrand becomes eˣ[xˣ + xˣ(1 + ln x)] = eˣ[f(x) + f'(x)] with f(x) = xˣ.
∫ eˣ[f(x) + f'(x)]dx = eˣ f(x) + C
- Apply the limits. [eˣ xˣ]₁² = e² × 2² - e¹ × 1¹ = 4e² - e.
- So. The integral equals 4e² - e = e(4e - 1), which is option (c).
The idea to keep. Whenever you see eˣ times a messy expression, try to write it as f(x) + f'(x). The antiderivative is then eˣ f(x), with no further work.
Answer from the NTA final answer key. Solution by PrepWiser.