CUET UG 2025 — Mathematics Calculus

The integral to evaluate is $\displaystyle \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}dx$∫−11​x2+2∣x∣+1x3+∣x∣+1​dx

The integrand contains absolute values and the limits cross $x = 0$x=0.

---

The denominator can be simplified:

$$x^2 + 2|x| + 1 = (|x|)^2 + 2|x| + 1$$x2+2∣x∣+1=(∣x∣)2+2∣x∣+1

$$= (|x| + 1)^2$$=(∣x∣+1)2

---

Let $f(x) = \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}$f(x)=x2+2∣x∣+1x3+∣x∣+1​

For $x > 0$x>0:

$$f(x) = \frac{x^3 + x + 1}{(x+1)^2}$$f(x)=(x+1)2x3+x+1​

$$f(-x) = \frac{-x^3 + x + 1}{(x+1)^2}$$f(−x)=(x+1)2−x3+x+1​

Adding these:

$$f(x) + f(-x) = \frac{(x^3 + x + 1) + (-x^3 + x + 1)}{(x+1)^2}$$f(x)+f(−x)=(x+1)2(x3+x+1)+(−x3+x+1)​

$$= \frac{2x + 2}{(x+1)^2}$$=(x+1)22x+2​

$$= \frac{2(x+1)}{(x+1)^2}$$=(x+1)22(x+1)​

$$= \frac{2}{x+1}$$=x+12​

---

Using the symmetry property for integrals over $[-a, a]$[−a,a]:

$$\int_{-1}^1 f(x)dx = \int_0^1 [f(x) + f(-x)]dx$$∫−11​f(x)dx=∫01​[f(x)+f(−x)]dx

Therefore:

$$\int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}dx = \int_0^1 \frac{2}{x+1}dx$$∫−11​x2+2∣x∣+1x3+∣x∣+1​dx=∫01​x+12​dx

---

Evaluating the integral:

$$\int_0^1 \frac{2}{x+1}dx = 2\int_0^1 \frac{1}{x+1}dx$$∫01​x+12​dx=2∫01​x+11​dx

$$= 2[\ln(x+1)]_0^1$$=2[ln(x+1)]01​

$$= 2[\ln(2) - \ln(1)]$$=2[ln(2)−ln(1)]

$$= 2\ln(2)$$=2ln(2)

$$= 2\log_e 2$$=2loge​2

Therefore, the answer is $2\log_e 2$2loge​2.