Indefinite IntegrationeasyFree

Indefinite Integration: Question

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

Question
If ∫x4+1x6+1 dx=αarctan⁡P(x)Q(x)+C\displaystyle\int\frac{x^4+1}{x^6+1}\,dx = \alpha\arctan\frac{P\left(x\right)}{Q\left(x\right)}+C, α∈Q\alpha \in \mathbb{Q}, then
Aα=13\alpha = \dfrac13correct
BP(1)=0P\left(1\right) = 0correct
CP(−1)=0P\left(-1\right) = 0correct
DQ(1)=−4Q\left(1\right) = -4correct
Solution
Step 1:
x6+1=(x2+1)(x4−x2+1)x^6+1 = \left(x^2+1\right)\left(x^4-x^2+1\right)
 and
x4+1=(x4−x2+1)+x2x^4+1 = \left(x^4-x^2+1\right)+x^2
x4+1x6+1=x4−x2+1(x2+1)(x4−x2+1)+x2x6+1=1x2+1+x2(x3)2+1\frac{x^4+1}{x^6+1} = \frac{x^4-x^2+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2}{x^6+1} = \frac{1}{x^2+1}+\frac{x^2}{\left(x^3\right)^2+1}
 Step 2:
∫dxx2+1=arctan⁡x,∫x2 dx(x3)2+1=13∫d(x3)(x3)2+1=13arctan⁡x3\int\frac{dx}{x^2+1} = \arctan x ,\qquad \int\frac{x^2\,dx}{\left(x^3\right)^2+1} = \frac13\int\frac{d\left(x^3\right)}{\left(x^3\right)^2+1} = \frac13\arctan x^3
∫x4+1x6+1 dx=arctan⁡x+13arctan⁡x3+C=13(3arctan⁡x+arctan⁡x3)+C\int\frac{x^4+1}{x^6+1}\,dx = \arctan x+\frac13\arctan x^3+C = \frac13\left(3\arctan x+\arctan x^3\right)+C
 ∴ α=13\therefore\ \alpha = \dfrac13, so **(A)** holds. Step 3: Put u=3arctan⁡xu = 3\arctan x and v=arctan⁡x3v = \arctan x^3.
tan⁡u=tan⁡(3arctan⁡x)=3x−x31−3x2(triple-angle formula)\tan u = \tan\left(3\arctan x\right) = \frac{3x-x^3}{1-3x^2}\qquad\text{(triple-angle formula)}
tan⁡(u+v)=tan⁡u+tan⁡v1−tan⁡utan⁡v=3x−x31−3x2+x31−3x−x31−3x2⋅x3\tan\left(u+v\right) = \frac{\tan u+\tan v}{1-\tan u\tan v} = \frac{\dfrac{3x-x^3}{1-3x^2}+x^3}{1-\dfrac{3x-x^3}{1-3x^2}\cdot x^3}
 Step 4: Numerator: 3x−x3+x3(1−3x2)1−3x2=3x−3x51−3x2\dfrac{3x-x^3+x^3\left(1-3x^2\right)}{1-3x^2} = \dfrac{3x-3x^5}{1-3x^2}. Denominator: (1−3x2)−x3(3x−x3)1−3x2=1−3x2−3x4+x61−3x2\dfrac{\left(1-3x^2\right)-x^3\left(3x-x^3\right)}{1-3x^2} = \dfrac{1-3x^2-3x^4+x^6}{1-3x^2}. On dividing,
tan⁡(u+v)=3x−3x51−3x2−3x4+x6\tan\left(u+v\right) = \frac{3x-3x^5}{1-3x^2-3x^4+x^6}
P(x)=3x−3x5,Q(x)=1−3x2−3x4+x6P\left(x\right) = 3x-3x^5 ,\qquad Q\left(x\right) = 1-3x^2-3x^4+x^6
∫x4+1x6+1 dx=13arctan⁡3x−3x51−3x2−3x4+x6+C\int\frac{x^4+1}{x^6+1}\,dx = \frac13\arctan\frac{3x-3x^5}{1-3x^2-3x^4+x^6}+C
 Step 5:
P(1)=3−3=0,P(−1)=−3+3=0P\left(1\right) = 3-3 = 0 ,\qquad P\left(-1\right) = -3+3 = 0
Q(1)=1−3−3+1=−4Q\left(1\right) = 1-3-3+1 = -4
 So all four statements hold. Differentiating the closed form gives back x4+1x6+1\dfrac{x^4+1}{x^6+1} exactly, which confirms this pair (P,Q)\left(P,Q\right). Answer: (1), (3), (3) and (4)
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