StemCET Logo

Maths Question 20 – JEE-MAIN 2026

← Back
The value of the integral ∫−11x3+|x|+1x2+2|x|+1dx is equal to :

For an integral with symmetric limits, consider using the property ∫−aaf(x)dx=∫0a(f(x)+f(−x))dx.

● Video Walkthrough
Step 1: Utilize Symmetry and Absolute Value Definition✦ Active

The given integral is I=∫−11x3+|x|+1x2+2|x|+1dx. Due to the symmetric limits, we can use the property ∫−aaf(x)dx=∫0a(f(x)+f(−x))dx. Let f(x)=x3+|x|+1x2+2|x|+1. Then, we find f(−x):

f(−x)=(−x)3+|−x|+1(−x)2+2|−x|+1=−x3+|x|+1x2+2|x|+1

Now, sum f(x) and f(−x):

f(x)+f(−x)=x3+|x|+1x2+2|x|+1+−x3+|x|+1x2+2|x|+1=2|x|+2x2+2|x|+1

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

f(x)+f(−x)=2(|x|+1)(|x|+1)2=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=∫012|x|+1dx=∫012x+1dx
💡 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|]01

Apply the limits of integration:

I=2(ln⁡(1+1)−ln⁡(0+1))

Since ln⁡1=0:

I=2(ln⁡2−0)=2ln⁡2

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

💡 Teacher's Secret Hint

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

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.