The integral $\int \frac{\left(x^{8}-x^{2}\right) d x}{\left(x^{12}+3 x^{6}+1\right) \tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)}$ is equal to :

(A) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{1 / 3}+C$

(B) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{1 / 2}+C$

(C) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)+C$

(D) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{3}+C$

Indefinite Integrals — JEE Main · 2024 · 27 Jan 2024 · Shift 2 Question with Solution

The integral

\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}

is equal to :

Answer: (A)

I = \int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) tan ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )} .

Let

t = tan^{- 1} (x^{3} + \frac{1}{x ^{3}})

. Then

d t = \frac{1}{1 + ( x ^{3} + \frac{1}{x ^{3}} ) ^{2}} (3 x^{2} - \frac{3}{x ^{4}}) d x .

Since

(x^{3} + \frac{1}{x ^{3}})^{2} = x^{6} + 2 + \frac{1}{x ^{6}}

, adding 1 gives

1 + (x^{3} + \frac{1}{x ^{3}})^{2} = \frac{x ^{12} + 3 x ^{6} + 1}{x ^{6}}

. So

d t = \frac{x ^{6}}{x ^{12} + 3 x ^{6} + 1} \cdot 3 (x^{2} - \frac{1}{x ^{4}}) d x = \frac{3 ( x ^{8} - x ^{2} )}{x ^{12} + 3 x ^{6} + 1} d x .

The integral becomes

I = \frac{1}{3} \int \frac{d t}{t} = \frac{1}{3} ln ∣ t ∣ + C = lo g_{e} (tan^{- 1} (x^{3} + \frac{1}{x ^{3}}))^{1/3} + C .

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option