CUET UG 2025 — Mathematics Calculus

The integral $\displaystyle I = \int \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{3\log_e x} - e^{2\log_e x}} dx$I=∫e3loge​x−e2loge​xe5loge​x−e4loge​x​dx can be simplified using the property $e^{a\log_e x} = x^a$ealoge​x=xa.

Applying this property to each term:

$$e^{5\log_e x} = x^5$$e5loge​x=x5

$$e^{4\log_e x} = x^4$$e4loge​x=x4

$$e^{3\log_e x} = x^3$$e3loge​x=x3

$$e^{2\log_e x} = x^2$$e2loge​x=x2

The integral becomes:

$$I = \int \frac{x^5 - x^4}{x^3 - x^2} dx$$I=∫x3−x2x5−x4​dx

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Factoring the numerator and denominator:

Numerator: $x^5 - x^4 = x^4(x - 1)$x5−x4=x4(x−1)

Denominator: $x^3 - x^2 = x^2(x - 1)$x3−x2=x2(x−1)

$$I = \int \frac{x^4(x - 1)}{x^2(x - 1)} dx$$I=∫x2(x−1)x4(x−1)​dx

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Canceling $(x-1)$(x−1) from numerator and denominator, and simplifying:

$$I = \int \frac{x^4}{x^2} dx$$I=∫x2x4​dx

$$I = \int x^2 dx$$I=∫x2dx

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Using the power rule for integration:

$$I = \frac{x^{3}}{3} + C$$I=3x3​+C

Therefore, the integral equals $\frac{x^3}{3} + C$3x3​+C.