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Divide by x^2 and substitute t = x + 1/x (denominators like x^4 + kx^2 + 1)

This type of question, asked 4 times: 16 May 2023 S2, 17 Apr 2026 S1, 20 Apr 2026 S1, 12 May 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    16 May 2023 S2

    113
    ∫x2−4x4+9x2+16 dx=tan⁡−1(f(x))+c\int \frac{x^2-4}{x^4+9x^2+16}\,dx = \tan^{-1}(f(x)) + c (where cc is a constant of integration), then value of f(2)f(2) is

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  2. Question 2

    17 Apr 2026 S1

    79
    The value of ∫sin⁡3x(cos⁡4x+3cos⁡2x+1)tan⁡−1(sec⁡x+cos⁡x) dx\displaystyle\int\dfrac{\sin^3 x}{(\cos^4 x + 3\cos^2 x + 1)\tan^{-1}(\sec x + \cos x)}\,dx is

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  3. Question 3

    20 Apr 2026 S1

    78
    ∫x4(x10−1)x20+3x10+1 dx=⋯\displaystyle\int\dfrac{x^4(x^{10} - 1)}{x^{20} + 3x^{10} + 1}\,dx = \cdots

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  4. Question 4

    12 May 2026 S1

    77
    For x≠0x \neq 0, the value of I=∫x2−1(x4+3x2+1)tan⁡−1(x+1x) dxI = \displaystyle\int\dfrac{x^2 - 1}{(x^4 + 3x^2 + 1)\tan^{-1}\left(x + \dfrac{1}{x}\right)}\,dx is ...

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