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CBSE Class 12 Math 2025 All Sets Solved Paper

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Question : 3 of 20
Marks: +1, -0
If ∫01ex1+x dx=α,\int\limits_{0}^{1} \frac{e^{x}}{1+x} \, dx = \alpha, then ∫01ex(1+x)2 dx\int\limits_{0}^{1} \frac{e^{x}}{(1+x)^{2}} \, dx is equal to
Solution:  
∫01ex1+x dx=α\int\limits_{0}^{1} \frac{e^{x}}{1+x} \, dx = \alpha
Here we will apply integration by parts formula
Which is ∫uv dx=u∫v dx−∫(u′∫v dx)dx\int u v \, dx = u \int v \, dx - \int \left( u' \int v \, dx \right) dx
Here u=11+xu = \frac{1}{1+x} and v=exv = e^{x}
⇒u′=−1(1+x)2\Rightarrow u' = \frac{-1}{(1+x)^{2}}
∫01ex1+xdx=[11+xex]01−∫01(−1(1+x)2)exdx\int\limits_{0}^{1} \frac{e^{x}}{1+x} dx = \left[ \frac{1}{1+x} e^{x} \right]_{0}^{1} - \int\limits_{0}^{1} \left( \frac{-1}{(1+x)^{2}} \right) e^{x} dx
α=e2−1+∫01ex1+x2dx\alpha = \frac{e}{2} - 1 + \int\limits_{0}^{1} \frac{e^{x}}{1+x^{2}} dx
⇒∫01ex(1+x)2dx=α+1−e2\Rightarrow \int\limits_{0}^{1} \frac{e^{x}}{(1+x)^{2}} dx = \alpha + 1 - \frac{e}{2}
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