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It is given that $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2).$$ (i) Find A and B so that $$\frac{16}{x^4 + 4} = \frac{A + 2x}{x^2 + 2x + 2} + \frac{B - 2x}{x^2 - 2x + 2}.$$ (ii) Hence, or otherwise, show that for any real number m, $$\int_{0}^{m} \frac{16}{x^4 + 4} \, dx = \ln \left( \frac{m^2 + 2m + 2}{m^2 - 2m + 2} \right) + 2 \tan^{-1}(m + 1) + 2 \tan^{-1}(m - 1).$$ (iii) Find the limiting value as m → ∞ of $$\int_{0}^{m} \frac{16}{x^4 + 4} \, dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2017 - Paper 1

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Question 14

It-is-given-that-$$x^4-+-4-=-(x^2-+-2x-+-2)(x^2---2x-+-2).$$--(i)-Find-A-and-B-so-that-$$\frac{16}{x^4-+-4}-=-\frac{A-+-2x}{x^2-+-2x-+-2}-+-\frac{B---2x}{x^2---2x-+-2}.$$--(ii)-Hence,-or-otherwise,-show-that-for-any-real-number-m,-$$\int_{0}^{m}-\frac{16}{x^4-+-4}-\,-dx-=-\ln-\left(-\frac{m^2-+-2m-+-2}{m^2---2m-+-2}-\right)-+-2-\tan^{-1}(m-+-1)-+-2-\tan^{-1}(m---1).$$--(iii)-Find-the-limiting-value-as-m-→-∞-of-$$\int_{0}^{m}-\frac{16}{x^4-+-4}-\,-dx.$$-HSC-SSCE Mathematics Extension 2-Question 14-2017-Paper 1.png

It is given that $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2).$$ (i) Find A and B so that $$\frac{16}{x^4 + 4} = \frac{A + 2x}{x^2 + 2x + 2} + \frac{B - 2x}{x^2 - 2x + ... show full transcript

Worked Solution & Example Answer:It is given that $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2).$$ (i) Find A and B so that $$\frac{16}{x^4 + 4} = \frac{A + 2x}{x^2 + 2x + 2} + \frac{B - 2x}{x^2 - 2x + 2}.$$ (ii) Hence, or otherwise, show that for any real number m, $$\int_{0}^{m} \frac{16}{x^4 + 4} \, dx = \ln \left( \frac{m^2 + 2m + 2}{m^2 - 2m + 2} \right) + 2 \tan^{-1}(m + 1) + 2 \tan^{-1}(m - 1).$$ (iii) Find the limiting value as m → ∞ of $$\int_{0}^{m} \frac{16}{x^4 + 4} \, dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2017 - Paper 1

Step 1

Find A and B so that $$\frac{16}{x^4 + 4} = \frac{A + 2x}{x^2 + 2x + 2} + \frac{B - 2x}{x^2 - 2x + 2}.$$

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Answer

To find A and B, we first multiply through by the denominator, yielding:

16=(A+2x)(x2−2x+2)+(B−2x)(x2+2x+2).16 = (A + 2x)(x^2 - 2x + 2) + (B - 2x)(x^2 + 2x + 2).

Expanding both sides:

  1. Expanding (A+2x)(x2−2x+2)(A + 2x)(x^2 - 2x + 2):

    • Ax2−2Ax+2A+2x3−4x2+4xAx^2 - 2Ax + 2A + 2x^3 - 4x^2 + 4x = 2x3+(A−4)x2+(4+2A)x2x^3 + (A - 4)x^2 + (4 + 2A)x

  2. Expanding (B−2x)(x2+2x+2)(B - 2x)(x^2 + 2x + 2):

    • Bx2+2Bx+2B−2x3−4x2−4xBx^2 + 2Bx + 2B - 2x^3 - 4x^2 - 4x = −2x3+(B−4)x2+(2B−4)x-2x^3 + (B - 4)x^2 + (2B - 4)x

Combining:

extTotal:(2−2)x3+(A+B−4)x2+(4+2A+2B−4)x+(2A+2B)=16 ext{Total: } (2 - 2)x^3 + (A + B - 4)x^2 + (4 + 2A + 2B - 4)x + (2A + 2B) = 16

Thus we have:

  • Coefficient of x3x^3: 2−2=02 - 2 = 0
  • Coefficient of x2x^2: A+B−4=0A + B - 4 = 0
  • Coefficient of xx: 4+2A+2B−4=04 + 2A + 2B - 4 = 0
  • Constant: 2A+2B=162A + 2B = 16

From the first equation, A+B=4A + B = 4. Substituting B=4−AB = 4 - A into the second:

2A+2(4−A)=162A + 2(4 - A) = 16 2A+8−2A=162A + 8 - 2A = 16 8=16ext(trivial,hence,norestrictions)8 = 16 ext{ (trivial, hence, no restrictions)}

From A+B=4A + B = 4: Set A=8,B=−4A = 8, B = -4. Thus: A=8,B=0.A = 8, B = 0.

Step 2

Hence, or otherwise, show that for any real number m, $$\int_{0}^{m} \frac{16}{x^4 + 4} \, dx = \ln \left( \frac{m^2 + 2m + 2}{m^2 - 2m + 2} \right) + 2 \tan^{-1}(m + 1) + 2 \tan^{-1}(m - 1).$$

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Answer

Starting from:

∫16x4+4 dx=∫(8x2+2x+2+8x2−2x+2) dx.\int \frac{16}{x^4 + 4} \, dx = \int \left( \frac{8}{x^2 + 2x + 2} + \frac{8}{x^2 - 2x + 2} \right) \, dx.

Calculating each integral:

  1. For 8x2+2x+2\frac{8}{x^2 + 2x + 2}:

    • Completing the square: 8(x+1)2+1\frac{8}{(x+1)^2 + 1} gives:
    • 8tan⁡−1(x+1).8 \tan^{-1}(x + 1).

  2. For 8x2−2x+2\frac{8}{x^2 - 2x + 2}:

    • Completing the square: 8(x−1)2+1\frac{8}{(x-1)^2 + 1} gives:
    • 8tan⁡−1(x−1).8 \tan^{-1}(x - 1).

Putting it all together:

∫0m16x4+4 dx=8tan⁡−1(m+1)+8tan⁡−1(m−1)+C. \int_0^m \frac{16}{x^4 + 4} \, dx = 8 \tan^{-1}(m + 1) + 8 \tan^{-1}(m - 1) + C.

Evaluating gives:

  • 16ln⁡(m2+2m+2m2−2m+2).16 \ln \left( \frac{m^2 + 2m + 2}{m^2 - 2m + 2} \right).

Step 3

Find the limiting value as m → ∞ of $$\int_{0}^{m} \frac{16}{x^4 + 4} \, dx.$$

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Answer

As mo∞m o \infty:

lim⁡m→∞∫0m16x4+4 dx=lim⁡m→∞(ln⁡(m2+2m+2m2−2m+2)+2tan⁡−1(m+1)+2tan⁡−1(m−1)).\lim_{m \to \infty} \int_0^m \frac{16}{x^4 + 4} \, dx = \lim_{m \to \infty} \left( \ln \left( \frac{m^2 + 2m + 2}{m^2 - 2m + 2} \right) + 2\tan^{-1}(m + 1) + 2\tan^{-1}(m - 1) \right).

Observing each term:

  • As m→∞m \to \infty, \tan^{-1}(m) approaches π2\frac{\pi}{2}.
  • Thus: 2tan⁡−1(m+1)+2tan⁡−1(m−1)→π.2\tan^{-1}(m + 1) + 2\tan^{-1}(m - 1) \to \pi.
  • The natural logarithm approaches 0 as the dominators grow without bound.

Thus, the limiting value is:

lim⁡m→∞∫0m16x4+4 dx=π.\lim_{m \to \infty} \int_0^m \frac{16}{x^4 + 4} \, dx = \pi.

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