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Integral of (x^9 - x^5 + x)/(3x^8 - 4x^4 + 6)^(3/4) from 0 to 1 | JEE Advanced

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

Question
Let I=∫01(x9−x5+x)(3x8−4x4+6)3/4dxI = \displaystyle\int_0^1\frac{\left(x^9-x^5+x\right)}{\left(3x^8-4x^4+6\right)^{3/4}}dx, then the value of (12 I)4\left(12\,I\right)^4 is
Solution
Answer: 5
Step 1: 3x8−4x4+6=3(x4−23)2+143>03x^8-4x^4+6 = 3\left(x^4-\tfrac23\right)^2+\tfrac{14}3>0 for every xx, so the 34\tfrac34 power is defined throughout. Since (x8)3/4=x6\left(x^8\right)^{3/4} = x^6, multiplying numerator and denominator by x6x^6 pulls a factor x8x^8 inside the bracket:
I=∫01(x9−x5+x)x6(3x8−4x4+6)3/4(x8)3/4dx=∫01x15−x11+x7(3x16−4x12+6x8)3/4dxI = \int_0^1\frac{\left(x^9-x^5+x\right)x^6} {\left(3x^8-4x^4+6\right)^{3/4}\left(x^8\right)^{3/4}}dx = \int_0^1\frac{x^{15}-x^{11}+x^7}{\left(3x^{16}-4x^{12}+6x^8\right)^{3/4}}dx
 Step 2: The numerator is proportional to the derivative of the bracket. Put
t=3x16−4x12+6x8⟹dtdx=48x15−48x11+48x7=48(x15−x11+x7)t = 3x^{16}-4x^{12}+6x^8 \quad\Longrightarrow\quad \frac{dt}{dx} = 48x^{15}-48x^{11}+48x^7 = 48\left(x^{15}-x^{11}+x^7\right)
 so (x15−x11+x7)dx=dt48\left(x^{15}-x^{11}+x^7\right)dx = \dfrac{dt}{48}, and
x=0⇒t=0,x=1⇒t=3−4+6=5x = 0 \Rightarrow t = 0,\qquad x = 1 \Rightarrow t = 3-4+6 = 5
 Step 3:
I=148∫05t−3/4dt=148[4t1/4]05=4⋅51/448=51/412I = \frac1{48}\int_0^5 t^{-3/4}dt = \frac1{48}\Big[4t^{1/4}\Big]_0^5 = \frac{4\cdot5^{1/4}}{48} = \frac{5^{1/4}}{12}
 The integrand is improper only at t=0t = 0, where t−3/4t^{-3/4} is integrable, so the value is finite. Answer: 55
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