CUET UG 2025 — Mathematics Calculus

For $x > 1$x>1, the integral can be simplified using the property $e^{a\log x} = x^a$ealogx=xa.

Applying this property:

$$e^{7\log x} = x^7$$e7logx=x7

$$e^{5\log x} = x^5$$e5logx=x5

$$e^{4\log x} = x^4$$e4logx=x4

The integral becomes:

$$\int \frac{x^7 - x^5}{x^5 - x^4} dx$$∫x5−x4x7−x5​dx

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Factoring the numerator and denominator:

$$x^7 - x^5 = x^5(x^2 - 1)$$x7−x5=x5(x2−1)

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

The integral becomes:

$$\int \frac{x^5(x^2 - 1)}{x^4(x - 1)} dx$$∫x4(x−1)x5(x2−1)​dx

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

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The numerator can be factored as a difference of squares:

$$x^2 - 1 = (x+1)(x-1)$$x2−1=(x+1)(x−1)

Substituting:

$$\int \frac{x(x+1)(x-1)}{x-1} dx$$∫x−1x(x+1)(x−1)​dx

Since $x > 1$x>1, $(x-1) \neq 0$(x−1)=0:

$$= \int x(x+1) dx$$=∫x(x+1)dx

$$= \int (x^2 + x) dx$$=∫(x2+x)dx

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Integrating term by term:

$$\int (x^2 + x) dx = \frac{x^3}{3} + \frac{x^2}{2} + C$$∫(x2+x)dx=3x3​+2x2​+C

Therefore, the integral equals $\dfrac{x^3}{3} + \dfrac{x^2}{2} + C$3x3​+2x2​+C where $C$C is a constant of integration.