CUET UG 2025 — Mathematics Calculus

Look at the integral: $\displaystyle \int_1^3 \frac{x^2}{x^3+1}dx$∫13​x3+1x2​dx

The denominator is $x^3 + 1$x3+1 and the numerator is $x^2$x2. The derivative of $x^3 + 1$x3+1 is $3x^2$3x2, which is almost what appears in the numerator.

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Let $u = x^3 + 1$u=x3+1

$$\frac{du}{dx} = 3x^2$$dxdu​=3x2

$$du = 3x^2 \, dx$$du=3x2dx

$$x^2 \, dx = \frac{1}{3} du$$x2dx=31​du

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The limits change from $x$x to $u$u:

When $x = 1$x=1: $u = (1)^3 + 1 = 2$u=(1)3+1=2

When $x = 3$x=3: $u = (3)^3 + 1 = 28$u=(3)3+1=28

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$$\int_1^3 \frac{x^2}{x^3+1}dx = \int_2^{28} \frac{1}{u} \cdot \frac{1}{3} du$$∫13​x3+1x2​dx=∫228​u1​⋅31​du

$$= \frac{1}{3} \int_2^{28} \frac{1}{u} \, du$$=31​∫228​u1​du

$$= \frac{1}{3} [\log u]_2^{28}$$=31​[logu]228​

$$= \frac{1}{3} (\log 28 - \log 2)$$=31​(log28−log2)

$$= \frac{1}{3} \log \frac{28}{2}$$=31​log228​

$$= \frac{1}{3} \log 14$$=31​log14

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Therefore, $\displaystyle \int_1^3 \frac{x^2}{x^3+1}dx = \frac{1}{3}\log 14$∫13​x3+1x2​dx=31​log14