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Definite Integration: Value Integral

JEE Maths question with a full step-by-step solution.

Question
The value of the integral
∫−11dx(x2+x+1)+x4+3x2+1\int_{-1}^{1}\frac{dx}{\left(x^2+x+1\right)+\sqrt{x^4+3x^2+1}}
 is
A122\dfrac{1}{2\sqrt2}
B00
Cπ2\dfrac\pi2
Dπ4\dfrac\pi4correct
Solution
Step 1: The limits are symmetric, so pair xx with −x-x. Writing f(x)f\left(x\right) for the integrand,
∫−11f(x)dx=∫01[f(x)+f(−x)]dx\int_{-1}^{1}f\left(x\right)dx = \int_0^1\left[f\left(x\right)+f\left(-x\right)\right]dx
 x4+3x2+1x^4+3x^2+1 has only even powers, so the square root is even and only the xx in x2+x+1x^2+x+1 changes sign. Step 2: Let
A=x2+1,S=x4+3x2+1A = x^2+1 ,\qquad S = \sqrt{x^4+3x^2+1}
 so
f(x)=1A+x+S,f(−x)=1A−x+Sf\left(x\right) = \frac{1}{A+x+S},\qquad f\left(-x\right) = \frac{1}{A-x+S}
 Step 3:
f(x)+f(−x)=(A−x+S)+(A+x+S)(A+S)2−x2=2(A+S)(A+S)2−x2f\left(x\right)+f\left(-x\right) = \frac{\left(A-x+S\right)+\left(A+x+S\right)}{\left(A+S\right)^2-x^2} = \frac{2\left(A+S\right)}{\left(A+S\right)^2-x^2}
 Step 4:
(A+S)2−x2=A2+2AS+S2−x2\left(A+S\right)^2-x^2 = A^2+2AS+S^2-x^2
A2+S2=(x4+2x2+1)+(x4+3x2+1)=2x4+5x2+2A^2+S^2 = \left(x^4+2x^2+1\right)+\left(x^4+3x^2+1\right) = 2x^4+5x^2+2
A2+S2−x2=2x4+4x2+2=2(x2+1)2=2A2A^2+S^2-x^2 = 2x^4+4x^2+2 = 2\left(x^2+1\right)^2 = 2A^2
(A+S)2−x2=2A2+2AS=2A(A+S)\left(A+S\right)^2-x^2 = 2A^2+2AS = 2A\left(A+S\right)
 Step 5: A=x2+1≥1A = x^2+1 \ge 1 and x4+3x2+1≥1x^4+3x^2+1 \ge 1 gives S≥1S \ge 1, so A+S≥2≠0A+S \ge 2 \ne 0 and it may be cancelled:
f(x)+f(−x)=2(A+S)2A(A+S)=1A=1x2+1f\left(x\right)+f\left(-x\right) = \frac{2\left(A+S\right)}{2A\left(A+S\right)} = \frac1A = \frac{1}{x^2+1}
 The square root vanished. Step 6:
∫−11f(x)dx=∫01dx1+x2=[tan⁡−1x]01=π4\int_{-1}^{1}f\left(x\right)dx = \int_0^1\frac{dx}{1+x^2} = \left[\tan^{-1}x\right]_0^1 = \frac\pi4
 Answer: (4)
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