For real numbers - and - if- x2 - 1 - tan - 1 x2 - 1x x4 - 3x2 - 1tan - 1 x2 - 1xdx- - log e tan - 1 x2 - 1x - - tan - 1 - x2 - 1x - - tan - 1 x2 - 1x - C where C is an arbitrary constant- then the value of 10- is equal to

I=∫x^2 1( x^4 + 3x^2 + 1)tan^ 1( x^2 + 1x)dx + ∫1x^4 + 3x^2 + 1dx = nbsp; ∫1 1/x^2/[ ( x + 1/x)^2 + 1]tan^ 1( x + 1/x)dx + ∫dx/x^4 + 3x^2 + 1 nbsp; ...

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Byju's Answer

Standard X

Mathematics

Sum of Infinite Terms of a GP

For real numb...

Question

For real numbers $α, β, γ$ and $δ,$ if
$\int \frac{(x^{2} - 1) + {tan}^{- 1} (\frac{x^{2} + 1}{x})}{(x^{4} + 3 x^{2} + 1) {tan}^{- 1} (\frac{x^{2} + 1}{x})} d x = α {log}_{e} ({tan}^{- 1} (\frac{x^{2} + 1}{x})) + β {tan}^{- 1} (\frac{γ (x^{2} - 1)}{x}) + δ {tan}^{- 1} (\frac{x^{2} + 1}{x}) + C$
where $C$ is an arbitrary constant, then the value of $10 (α + β γ + δ)$ is equal to

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Solution

$I = \int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) {tan}^{- 1} (\frac{x^{2} + 1}{x})} d x + \int \frac{1}{x^{4} + 3 x^{2} + 1} d x$

$= \begin{matrix} \int \frac{1 - \frac{1}{x^{2}}}{[{(x + \frac{1}{x})}^{2} + 1] {tan}^{- 1} (x + \frac{1}{x})} d x + \int \frac{d x}{x^{4} + 3 x^{2} + 1} ↓ ↓ I_{1} I_{2} \end{matrix}$
Let ${tan}^{- 1} (x + \frac{1}{x}) = t$
Then $I_{1} = \int \frac{d t}{t}$
$\Rightarrow I_{1} = ln | t | = ln ∣ ∣ ∣ {tan}^{- 1} (x + \frac{1}{x}) ∣ ∣ ∣$

Now, $I_{2} = \int \frac{d x}{x^{4} + 3 x^{2} + 1}$
$= \frac{1}{2} \int \frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + 3 x^{2} + 1} d x$
$= \frac{1}{2} ⎡ ⎢ ⎢ ⎢ ⎣ \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + 3 + \frac{1}{x^{2}}} d x - \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + 3 + \frac{1}{x^{2}}} d x ⎤ ⎥ ⎥ ⎥ ⎦$
$= \begin{matrix} \frac{1}{2} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 5} - \int \frac{(1 - \frac{1}{x^{2}})}{{(x + \frac{1}{x})}^{2} + 1} d x ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ↓ ↓ x - \frac{1}{x} = u x + \frac{1}{x} = v \end{matrix}$
$= \frac{1}{2} ⎡ ⎢ ⎢ ⎢ ⎣ \int \frac{d u}{u^{2} + {(\sqrt{5})}^{2}} - \int \frac{d v}{v^{2} + 1} ⎤ ⎥ ⎥ ⎥ ⎦$
$I_{2} = \frac{1}{2 \sqrt{5}} {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{x - \frac{1}{x}}{\sqrt{5}} ⎞ ⎟ ⎟ ⎟ ⎠ - \frac{1}{2} {tan}^{- 1} (x + \frac{1}{x})$
$I = I_{1} + I_{2} = ln ∣ ∣ ∣ {tan}^{- 1} (x + \frac{1}{x}) ∣ ∣ ∣ + \frac{1}{2 \sqrt{5}} ln (\frac{x^{2} - 1}{\sqrt{5} x}) - \frac{1}{2} {tan}^{- 1} (\frac{x^{2} + 1}{x}) + C$

$∴ α = 1, β = \frac{1}{2 \sqrt{5}}, γ = \frac{1}{\sqrt{5}}, δ = - \frac{1}{2}$

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For real numbers α,β,γ and δ, if ∫(x2−1)+tan−1(x2+1x)(x4+3x2+1)tan−1(x2+1x)dx=α loge(tan−1(x2+1x))+β tan−1(γ(x2−1)x)+δ tan−1(x2+1x)+C where C is an arbitrary constant, then the value of 10(α+βγ+δ) is equal to