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Mathematics Question on Integration by Partial Fractions

The integral ∫x8−x2(x12+3x6+1)tan⁡−1(x3+1x3) dx\int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)} \, dx is equal to:

A

log⁡e[tan⁡−1(x3+1x3)]1/3+C\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C

B

log⁡e[tan⁡−1(x3+1x3)]1/2+C\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/2} + C

C

log⁡e[tan⁡−1(x3+1x3)]+C\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right] + C

D

log⁡e[tan⁡−1(x3+1x3)]3+C\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{3} + C

Answer

log⁡e[tan⁡−1(x3+1x3)]1/3+C\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C

Explanation

Solution

Given:
I=∫x8−x2(x12+3x6+1)tan⁡−1(x3+1x3)dx.I = \int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)} dx.

Let:
t=tan⁡−1(x3+1x3).t = \tan^{-1}\left(x^3 + \frac{1}{x^3}\right).

Then:
dt=11+(x3+1x3)2⋅(3x2−3x4)dx.dt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \left(3x^2 - \frac{3}{x^4}\right) dx.

Simplifying:
dt=11+(x3+1x3)2⋅3x6−3x4dx.dt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \frac{3x^6 - 3}{x^4} dx.

dt=x6−1x12+3x6+1dx.dt = \frac{x^6 - 1}{x^{12} + 3x^6 + 1} dx.

Rewriting the integral:
I=13∫dtt=13ln⁡∣t∣+C.I = \frac{1}{3} \int \frac{dt}{t} = \frac{1}{3} \ln|t| + C.

Substituting back:
I=13ln⁡∣tan⁡−1(x3+1x3)∣+C.I = \frac{1}{3} \ln\left|\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right| + C.

Simplifying further:
I=ln⁡(tan⁡−1(x3+1x3))1/3+C.I = \ln\left(\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right)^{1/3} + C.

The correct option is (A) : log⁡e[tan⁡−1(x3+1x3)]1/3+C\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C