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Definite Integration: Let Value

JEE Maths question with a full step-by-step solution.

Question
Let f(x)=∫x1ln⁡(1−t2)dtf(x) = \displaystyle\int_x^1\ln\left(1-t^2\right)dt, x∈(0,1)x \in \left(0,1\right) and α=∫01f(x) dx\alpha = \displaystyle\int_0^1f(x)\,dx, then the value of 4α+54\alpha+5 is
Solution
Answer: 3
Step 1: Write α\alpha as a double integral and swap the order:
α=∫01(∫x1ln⁡(1−t2)dt)dx\alpha = \int_0^1\left(\int_x^1\ln\left(1-t^2\right)dt\right)dx
 The region is {(x,t):0<x<t<1}\left\{\left(x,t\right):0<x<t<1\right\}. Integrating in xx first, for a fixed tt the variable xx runs from 00 to tt:
α=∫01ln⁡(1−t2)(∫0tdx)dt=∫01tln⁡(1−t2)dt\alpha = \int_0^1\ln\left(1-t^2\right)\left(\int_0^t dx\right)dt = \int_0^1 t\ln\left(1-t^2\right)dt
 Step 2: Putting u=t2u = t^2, du=2t dtdu = 2t\,dt,
α=12∫01ln⁡(1−u)du\alpha = \frac12\int_0^1\ln\left(1-u\right)du
 Step 3:
∫01ln⁡(1−u)du=[−(1−u)ln⁡(1−u)−u]01=0−1−0=−1\int_0^1\ln\left(1-u\right)du = \left[-\left(1-u\right)\ln\left(1-u\right)-u\right]_0^1 = 0-1-0 = -1
 the boundary term (1−u)ln⁡(1−u)→0\left(1-u\right)\ln\left(1-u\right) \to 0 as u→1−u \to 1^-. Hence
α=−12\alpha = -\frac12
 Step 4:
4α+5=4(−12)+5=−2+5=34\alpha+5 = 4\left(-\frac12\right)+5 = -2+5 = 3
 Answer: 33
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