PYQ Vault
EbooksTutor

Integral with inverse tan of (x + 1/x) in the denominator, by substituting that inverse tan

This type of question, asked 2 times: 2021 Paper 6, 27 Jan 2024. All 2 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 6

    84
    For real numbers α,β,γ\alpha,\beta,\gamma and δ\delta, if
    ∫(x2−1)+tan⁡−1(x2+1x)(x4+3x2+1)tan⁡−1(x2+1x)dx=αlog⁡e(tan⁡−1(x2+1x))\begin{gathered}\int\frac{\left( x^{2}- 1 \right)+\tan^{- 1}\left( \frac{x^{2}+ 1}{x} \right)}{\left( x^{4}+ 3x^{2}+ 1 \right)\tan^{- 1}\left( \frac{x^{2}+ 1}{x} \right)}dx \\ =\alpha\log_{e}\left( \tan^{- 1}\left( \frac{x^{2}+ 1}{x} \right) \right)\end{gathered}
    +βtan⁡−1(γ(x2−1)x)+δtan⁡−1(x2+1x)+C+\beta\tan^{- 1}\left( \frac{\gamma\left( x^{2}- 1 \right)}{x} \right)+\delta\tan^{- 1}\left( \frac{x^{2}+ 1}{x} \right)+ C
    where CC is an arbitrary constant, then the value of 10(α+βγ+δ)10(\alpha+\beta\gamma+\delta) is equal to

  2. Question 2

    27 Jan 2024

    164
    The integral ∫(x8−x2)dx(x12+3x6+1)tan⁡−1(x3+1x3)\int\frac{\left( x^{8}-x^{2} \right)dx}{\left( x^{12}+ 3x^{6}+ 1 \right)\tan^{- 1}\left( x^{3}+\frac{1}{x^{3}} \right)} is equal to :

    Tap an option to check your answer.