CUET UG 2025 — Mathematics Algebra

The integral $\displaystyle I = \int e^x\left(\frac{x - 1}{3x^2}\right) dx$I=∫ex(3x2x−1​)dx

Simplify the fraction:

$$\frac{x - 1}{3x^2}$$3x2x−1​

$$= \frac{1}{3} \cdot \frac{x - 1}{x^2}$$=31​⋅x2x−1​

$$= \frac{1}{3}\left(\frac{x}{x^2} - \frac{1}{x^2}\right)$$=31​(x2x​−x21​)

$$= \frac{1}{3}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$=31​(x1​−x21​)

The integral becomes:

$$I = \frac{1}{3}\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right) dx$$I=31​∫ex(x1​−x21​)dx

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For integrals of the form $\displaystyle \int e^x[f(x) + f'(x)]dx = e^x \cdot f(x) + C$∫ex[f(x)+f′(x)]dx=ex⋅f(x)+C

Let $f(x) = \frac{1}{x}$f(x)=x1​

Then $f'(x) = -\frac{1}{x^2}$f′(x)=−x21​

Therefore:

$$f(x) + f'(x) = \frac{1}{x} + \left(-\frac{1}{x^2}\right)$$f(x)+f′(x)=x1​+(−x21​)

$$= \frac{1}{x} - \frac{1}{x^2}$$=x1​−x21​

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Applying the formula:

$$I = \frac{1}{3}\int e^x\left[\frac{1}{x} - \frac{1}{x^2}\right] dx$$I=31​∫ex[x1​−x21​]dx

$$= \frac{1}{3} \cdot e^x \cdot \frac{1}{x} + C$$=31​⋅ex⋅x1​+C

$$= \frac{e^x}{3x} + C$$=3xex​+C

Therefore, $I = \frac{e^x}{3x} + C$I=3xex​+C where $C$C is the constant of integration.