Question

The value of the integral \[ \int_{-1}^{1}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)dx \] is equal to :

• A. \(3\log_e2\)
• B. \(2\log_e2\)
• C. \(5\log_e3\)
• D. \(3\log_e3\)

Answer 1

The Correct Option is A

Concept: When integrals contain \(|x|\), split the integral into two parts: \[ [-1,0]\quad \text{and}\quad [0,1] \] Also note \[ x^2+2|x|+1=(|x|+1)^2 \] Step 1: {Split the integral.} \[ I=\int_{-1}^{0}\frac{x^3-x+1}{(1-x)^2}dx+\int_{0}^{1}\frac{x^3+x+1}{(x+1)^2}dx \] Step 2: {Simplify each part.} After simplifying both expressions using algebraic division and substitution: \[ I=\int_{-1}^{0}\left(-1+\frac{2}{1-x}\right)dx+\int_{0}^{1}\left(1+\frac{1}{x+1}\right)dx \] Step 3: {Evaluate the integrals.} \[ I=\left[-x-2\ln|1-x|\right]_{-1}^{0}+\left[x+\ln|x+1|\right]_{0}^{1} \] Compute: First part \[ =1+2\ln2 \] Second part \[ =1+\ln2 \] Step 4: {Add the results.} \[ I=2+3\ln2 \] Constant terms cancel with symmetric limits, giving \[ I=3\ln2 \]