The value of the integral integral from -1 to 1 of (x^3 + |x| + 1) / (x^2 + 2|x| + 1) dx is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q20 – JEE-MAIN 2026

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Home Maths IntegralsQuestion #209

The value of the integral

\int_{- 1}^{1} \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} d x

is equal to :

Clue 1 of 3: Symmetry Property

For an integral with symmetric limits, consider using the property $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x$ .

🧠 Full Solution Path

Step 1: Utilize Symmetry and Absolute Value Definition✦ Active

The given integral is $I = \int_{- 1}^{1} \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} d x$ . Due to the symmetric limits, we can use the property $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x$ . Let $f (x) = \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}$ . Then, we find $f (- x)$ :

f (- x) = \frac{(- x)^{3} + | - x | + 1}{(- x)^{2} + 2 | - x | + 1} = \frac{- x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}

Now, sum $f (x)$ and $f (- x)$ :

f (x) + f (- x) = \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} + \frac{- x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} = \frac{2 | x | + 2}{x^{2} + 2 | x | + 1}

The denominator can be rewritten as $(| x | + 1)^{2}$ . So, the expression simplifies to:

f (x) + f (- x) = \frac{2 (| x | + 1)}{(| x | + 1)^{2}} = \frac{2}{| x | + 1}

Step 2: Simplify the Integral○ Expand

Substitute the simplified expression back into the integral. For the interval $[0, 1]$ , $| x | = x$ . Therefore, the integral becomes:

I = \int_{0}^{1} \frac{2}{| x | + 1} d x = \int_{0}^{1} \frac{2}{x + 1} d x

💡 Teacher's Secret Hint

Remember to correctly handle the absolute value function based on the integration interval.

Step 3: Evaluate the Definite Integral○ Expand

Now, evaluate the definite integral:

I = 2 {[\ln | x + 1 |]}_{0}^{1}

Apply the limits of integration:

I = 2 (\ln (1 + 1) - \ln (0 + 1))

Since $\ln 1 = 0$ :

I = 2 (\ln 2 - 0) = 2 \ln 2

The value of the integral is $2 \ln 2$ , which corresponds to option 2.

💡 Teacher's Secret Hint

Ensure correct evaluation of logarithmic terms at the limits, especially $\ln 1 = 0$ .

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Symmetry Property

🧠 Full Solution Path