For real numbers \(\alpha, \beta, \gamma\) and \(\delta\), if \(\int \frac{\left(x^{2}-1\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)} d x\) \(=\alpha \log _{\mathrm{e}}\left(\tan ^{-1}\left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}}\right)\right)+\beta \tan ^{-1}\left(\frac{\gamma\left(\mathrm{x}^{2}-1\right)}{\mathrm{x}}\right)+\delta \tan ^{-1}\left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}}\right)+\mathrm{C}\) where \(C\) is an arbitrary constant, then the value of \(10(\alpha+\beta \gamma+\delta)\) is equal to........

Indefinite Integrals — JEE Main · 2021 · 16 Mar 2021 · Shift 2 Question with Solution

For real numbers

α, β, γ

and

δ

, if

\int \frac{( x ^{2} - 1 ) + t a n ^{- 1} ( \frac{x ^{2} + 1}{x} )}{( x ^{4} + 3 x ^{2} + 1 ) t a n ^{- 1} ( \frac{x ^{2} + 1}{x} )} d x

= α lo g_{e} (tan^{- 1} (\frac{x ^{2} + 1}{x})) + β tan^{- 1} (\frac{γ ( x ^{2} - 1 )}{x}) + δ tan^{- 1} (\frac{x ^{2} + 1}{x}) + C

where

C

is an arbitrary constant, then the value of

10 (α + β γ + δ)

is equal to........

Answer: 6

Put

θ = tan^{- 1} \frac{x ^{2} + 1}{x}

. Since

1 + (\frac{x ^{2} + 1}{x})^{2} = \frac{x ^{4} + 3 x ^{2} + 1}{x ^{2}}

and

\frac{d}{d x} \frac{x ^{2} + 1}{x} = \frac{x ^{2} - 1}{x ^{2}}

d θ = \frac{x ^{2} - 1}{x ^{4} + 3 x ^{2} + 1} d x

so the integrand splits into

\frac{1}{θ} \frac{d θ}{d x}

and

\frac{1}{x ^{4} + 3 x ^{2} + 1}

, giving

\int \frac{d θ}{θ} + \int \frac{d x}{x ^{4} + 3 x ^{2} + 1} = lo g_{e} θ + \int \frac{d x}{x ^{4} + 3 x ^{2} + 1}

For the remaining integral divide by

x^{2}

, write

\frac{1}{x ^{2}} = \frac{1}{2} [(1 + \frac{1}{x ^{2}}) - (1 - \frac{1}{x ^{2}})]

and use

x^{2} + 3 + \frac{1}{x ^{2}} = (x - \frac{1}{x})^{2} + 5 = (x + \frac{1}{x})^{2} + 1

, which gives

\int \frac{d x}{x ^{4} + 3 x ^{2} + 1} = \frac{1}{2 5} tan^{- 1} \frac{x ^{2} - 1}{5 x} - \frac{1}{2} tan^{- 1} \frac{x ^{2} + 1}{x}

Comparing with the stated form,

α = 1

β = \frac{1}{2 5}

γ = \frac{1}{5}

and

δ = - \frac{1}{2}

, so

10 (α + β γ + δ) = 10 (1 + \frac{1}{10} - \frac{1}{2}) = 6

Thus

10 (α + β γ + δ) = 6

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer