Verified answer

Note that

∫(x = 1 to 2) ln(x²-3x+3) dx/[(x-1)(x-2)]

= ∫(x = 1 to 2) ln[(1/4) ((2x - 3)² + 3)] dx/[(x-1)(x-2)], completing the square

= ∫(w = -1 to 1) ln[(w²+3)/4] * (dw/2) / [((w+3)/2 - 1)((w+3)/2 - 2)], letting w = 2x - 3

= 2 ∫(w = -1 to 1) [ln(w²+3) - ln 4] dw/(w² - 1)

= 4 ∫(w = 0 to 1) [ln(w²+3) - ln 4] dw/(w² - 1), since the integrand is even

Now, we let w = (1 - t)/(1 + t), obtaining

4 ∫(t = 1 to 0) [ln(((1 - t)/(1 + t))² + 3) - ln 4] * (-2 dt/(1 + t)²) / (((1 - t)/(1 + t))² - 1)

= 8 ∫(t = 0 to 1) [ln((1 - t)²/(1 + t)² + 3) - ln 4] dt / ((1 - t)² - (1 + t)²)

= 8 ∫(t = 0 to 1) ln((1 + t + t²)/(1 + 2t + t²)) dt / (-4t)

= 2 ∫(t = 0 to 1) [ln(1 + 2t + t²) - ln(1 + t + t²)] dt/t

= 2(π²/6 - π²/9)

= π²/9.

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For the next to last step, it suffices to compute

F(a) = ∫(t = 0 to 1) ln(1 + 2at + t²) dt/t.

Differentiating under the integral sign (with respect to a):

F'(a) = ∫(t = 0 to 1) 2 dt/(1 + 2at + t²)

.........= ∫(t = 0 to 1) 2 dt/[(t + a)² + (1 - a²)], completing the square

.........= (2/√(1 - a²)) arctan((t + a)/√(1 - a²)) {for t = 0 to 1}

.........= (2/√(1 - a²)) arctan [(a + 1)/√(1 - a²)) - arctan(a/√(1 - a²))]

.........= (2/√(1 - a²)) arctan [((a + 1)/√(1 - a²)) - a/√(1 - a²)) / {1 + a(a + 1)/(1 - a²)}]

.........= (2/√(1 - a²)) arctan [√(1 - a²)/(1 + a)]

Hence,

F(a) = ∫ (2/√(1 - a²)) arctan [√(1 - a²)/(1 + a)] da

.......= ∫ (2/cos θ) arctan [cos θ/(1 + sin θ)] * (cos θ dθ), letting a = sin θ

.......= ∫ 2 arctan [cos θ/(1 + sin θ)] dθ

However, tan(a/2) = sin a/(1 + cos a) by the half angle identity.

Thus, tan(π/4 - θ/2) = cos θ/(1 + sin θ), by letting a = π/2 - θ.

So, we find that

F(a) = ∫ 2 arctan [tan(π/4 - θ/2)] dθ

........= ∫ 2(π/4 - θ/2) dθ

........= πθ/2 - θ²/2 + C.

........= π(arcsin a)/2 - (arcsin a)²/2 + C.

To find C, note that

F(1) = ∫(t = 0 to 1) ln(t² + 2t + 1) dt/t

.......= ∫(t = 0 to 1) ln((t+1)²) dt/t

.......= 2 ∫(t = 0 to 1) ln(t+1) dt/t, via power rule for logs

.......= 2 ∫(t = 0 to 1) [Σ(k = 1 to ∞) (-1)^(k-1) t^k/k] dt/t, via Maclaurin series for ln(t+1)

.......= 2 ∫(t = 0 to 1) [Σ(k = 1 to ∞) (-1)^(k-1) t^(k-1)/k] dt

.......= 2 Σ(k = 1 to ∞) (-1)^(k-1) t^k/k² {for t = 0 to 1}

.......= 2 Σ(k = 1 to ∞) (-1)^(k-1)/k².

Now, we rewrite the series:

F(1) = 2 (1 - 1/2² + 1/3² - 1/4² + 1/5² - 1/6² + ...)

.......= 2 [(1 + 1/2² + 1/3² + 1/4² + 1/5² + 1/6² + ...) - (2/2² + 2/4² + 2/6² + ...)]

.......= 2 [(1 + 1/2² + 1/3² + 1/4² + 1/5² + 1/6² + ...) - (2/2²) (1 + 1/2² + 1/3² + ...)]

.......= 2 (1 - 2/2²) (1 + 1/2² + 1/3² + ...)

.......= 2 (1 - 1/2) (π²/6), since Σ(k = 1 to ∞) 1/k² = π²/6

.......= π²/6.

Now, we can solve for C, by using F(a) = π(arcsin a)/2 - (arcsin a)²/2 + C with a = 1:

F(1) = π(arcsin 1)/2 - (arcsin 1)² + C

==> π²/6 = π(π/2)/2 - (π/2)²/2 + C

==> C = π²/24.

Thus, ∫(t = 0 to 1) ln(1 + 2at + t²) dt/t = π(arcsin a)/2 - (arcsin a)²/2 + π²/24.

So, ∫(t = 0 to 1) ln(1 + t + t²) dt/t

= π(arcsin(1/2))/2 - (arcsin(1/2))²/2 + 5π²/24, letting a = 1/2

= π(π/6)/2 - (π/6)²/2 + π²/24

= π²/9, as we used above.

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(Final answer was double checked on Wolfram Alpha.)

I hope this helps!

int_1^2 ln(x²-3x+3)/[(x-1)(x-2)] dx

int_1^2 ln[((x - (3/2))^2) + (3/4)]/[(x - 1)*(x - 2)] dx

u = x - (3/2); du = dx; u + (3/2) = x

int_(-1/2)^(1/2) ln[(u^2) + (3/4)]/[(u + (1/2))*(u - (1/2))] du

int_(-1/2)^(1/2) sum_(k = 1)^(infinity) [((u^2) - (1/4))^(k - 1) * ((-1)^(k+1))/k] du

We can simplify because it is an even function therefore

2 * int_(0)^(1/2) sum_(k = 1)^(infinity) [((u^2) - (1/4))^(k - 1) * ((-1)^(k+1))/k] du

int_(0)^(pi/2) sum_(k = 1)^infinity [((cos(t))^(2k - 1))* ((1/4)^(k-1))]/k dt

Let k = n + 1; then make n = k

int_(0)^(pi/2) sum_(k = 1)^infinity [((cos(t))^(2k + 1))* ((1/4)^k)]/(k + 1) dt

sum_(k = 0)^infinity {((1/4)^k)/(k + 1) * int_(0)^(pi/2) (cos(t))^(2k + 1) dt}

int_(0)^(pi/2) (cos(t))^(2k + 1) dt = (4^k)*((k!)^2)/(2k + 1)!

sum_(k = 0)^infinity [((1/4)^k)/(k + 1)]*[(4^k)*((k!)^2)/(2k + 1)!]

sum_(k = 0)^infinity ((k!)^2)/[(k + 1)*((2k + 1)!)] is the final answer

Apply Simpson's rule with 6 subdivisions. because this is a definite integral.

Because (x-1)(x-2) = 0 when x=1 or 2, I have used the limits 1.00001 and 2.00001

Lower limit 1.00001

Upper limit 2.00001

Number of intervals 6

Δx = (2.00001 - 1.00001) / 6 = 0.16667

number of intervals 6

The intervals are:

[a(0),a(1)][a(1),a(2)][a(2),a(3)][a(3),a(4)]

[a(4),a(5)][a(5),a(6)]

(1.00001,1.16668), (1.16668,1.33334), (1.33334,1.50001), (1.50001,1.66668),

(1.66668,1.83334), (1.83334,2.00001),

The midpoints of each interval are

1.08334, 1.25001, 1.41668, 1.58334,

1.75001, 1.91668,

Simpson's Sum =

f(1.00001) +2f(1.166677) + 2f(1.333343) + 2f(1.50001) + 2f(1.666677) + 2f(1.833343) +

f(2.00001) +

4f(1.083343) + 4f(1.25001) + 4f(1.416677) + 4f(1.583343) + 4f(1.75001) +

f(1.916677)

Simpson's Sum =

1.000005 + 2(1.076633) + 2(1.130917) + 2(1.150728) + 2(1.130913) + 2(1.076624) +

0.999995 +

4(1.040263) + 4(1.107413) + 4(1.145687) + 4(1.145685) + 4(1.107407) +

4(1.040254)

Simpson's sum = Δx/6 [f(a[0])+2Σ sum of end points+4Σ sum of midpoints+f(a[n])]

=(0.166667/6)[1.000005 +13.13163 +26.346836 +0.999995]

= (0.027778)(39.478466) = 1.09662406

Area = 1.09662