Board 2026 Class 12 Maths Q9 Solution: ∫ from −1 to 1 of x³/(x² + 2|x| + 1) dx

The question

The value of $\displaystyle\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx$ is

(A) $0$

(B) $\log 2$

(C) $2\log 2$

(D) $\frac{1}{2}\log 2$

Exam answer

Let $f(x) = \frac{x^3}{x^2 + 2|x| + 1}$ ⁠. Then

\begin{aligned} f(-x) &= \frac{(-x)^3}{(-x)^2 + 2\lvert -x\rvert + 1} \\ &= \frac{-x^3}{x^2 + 2|x| + 1} \\ &= -f(x) \end{aligned}

Why?

To test for odd or even, replace $x$ by $-x$ everywhere:

\begin{aligned} (-x)^3 &= -x^3 \\ (-x)^2 &= x^2 \\ \lvert -x\rvert &= |x| \end{aligned}

Only the numerator changes sign, so $f(-x) = -f(x)$ ⁠.

So $f$ is odd, and $\int_{-a}^{a} f(x)\,dx = 0$ ⁠.

Why?

The property used:

\int_{-a}^{a} f(x)\,dx = 0

whenever $f(-x) = -f(x)$ ⁠. The graph of an odd function is symmetric about the origin, so the area above the axis on one side cancels the equal area below it on the other side.

The limits must be equal and opposite. Here they are $-1$ and $1$ ⁠, so $a = 1$ ⁠. There is no need to integrate, which would mean splitting at 0 because of $|x|$ ⁠.

Here $a = 1$ ⁠, so the value is $0$ ⁠.

Answer: (A) $0$

Where this comes from

Chapter 7, Integrals: Definite integrals: odd function property.

Question type: Limits from $-a$ to $a$

How to spot it: click to see

Type 11. Limits from $-a$ to $a$ ⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Here: The limits are $-1$ and $1$ ⁠. Putting in $-x$ ⁠, only the top $x^3$ changes sign, since $x^2$ and $|x|$ do not, so the integrand is odd and the answer is 0 with no integrating.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Q9: $\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx$