CUET UG 2025 — Mathematics Calculus

Given $\displaystyle \int_0^1 \frac{e^x}{1 + x} dx = m$∫01​1+xex​dx=m

The two integrals are similar - the second has $(1+x)^2$(1+x)2 in the denominator instead of $(1+x)$(1+x). Integration by Parts can connect these forms.

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For $\displaystyle \int_0^1 \frac{e^x}{(1 + x)^2} dx$∫01​(1+x)2ex​dx, let:

$u = e^x$u=ex → $du = e^x dx$du=exdx

$dv = \frac{1}{(1+x)^2} dx$dv=(1+x)21​dx → $v = -\frac{1}{1+x}$v=−1+x1​

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Applying Integration by Parts:

$$\int_0^1 \frac{e^x}{(1 + x)^2} dx = \left[e^x \cdot \left(-\frac{1}{1+x}\right)\right]_0^1 - \int_0^1 \left(-\frac{1}{1+x}\right) \cdot e^x dx$$∫01​(1+x)2ex​dx=[ex⋅(−1+x1​)]01​−∫01​(−1+x1​)⋅exdx

$$= \left[-\frac{e^x}{1+x}\right]_0^1 + \int_0^1 \frac{e^x}{1+x} dx$$=[−1+xex​]01​+∫01​1+xex​dx

$$= \left[-\frac{e^x}{1+x}\right]_0^1 + m$$=[−1+xex​]01​+m

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Evaluating $\left[-\frac{e^x}{1+x}\right]_0^1$[−1+xex​]01​:

At $x = 1$x=1: $-\frac{e^1}{1+1} = -\frac{e}{2}$−1+1e1​=−2e​

At $x = 0$x=0: $-\frac{e^0}{1+0} = -1$−1+0e0​=−1

$$\left[-\frac{e^x}{1+x}\right]_0^1 = -\frac{e}{2} - (-1)$$[−1+xex​]01​=−2e​−(−1)

$$= -\frac{e}{2} + 1$$=−2e​+1

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$$\int_0^1 \frac{e^x}{(1 + x)^2} dx = -\frac{e}{2} + 1 + m$$∫01​(1+x)2ex​dx=−2e​+1+m

$$= m - \frac{e}{2} + 1$$=m−2e​+1

Therefore, the value is $m - \frac{e}{2} + 1$m−2e​+1