1.Answer: Divide by $x^2 \implies \int \frac{1-1/x^2}{x^2+1/x^2} dx$. Let $x+1/x = t, (1-1/x^2)dx = dt$. Integral is $\int \frac{dt}{t^2-2} = \frac{1}{2\sqrt{2}}\ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right| + C$.
2.Answer: Divide by $x^2 \implies \int \frac{1+1/x^2}{x^2+1+1/x^2} dx$. Let $x-1/x = t, dt=(1+1/x^2)dx$. Integral is $\int \frac{dt}{t^2+3} = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right) + C$.
3.Answer: $\int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx = \sqrt{2}\int \frac{\sin x + \cos x}{\sqrt{1-(\sin x - \cos x)^2}} dx$. Let $t=\sin x - \cos x \implies dt = (\cos x + \sin x)dx$. Integral is $\sqrt{2}\sin^{-1}(\sin x - \cos x) + C$.
4.Answer: Rewrite $1$ as $\frac{1}{2}[(x^2+1) - (x^2-1)]$. Splits into $\frac{1}{2}\int \frac{x^2+1}{x^4+1} dx - \frac{1}{2}\int \frac{x^2-1}{x^4+1} dx$. Using logic from Q1 & Q2: $\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right) - \frac{1}{4\sqrt{2}}\ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right| + C$.
5.Answer: Extract $x^4$ from the bracket: $\int \frac{dx}{x^5(1+x^{-4})^{3/4}}$. Let $1+x^{-4}=t \implies -4x^{-5}dx = dt$. Integral is $-\frac{1}{4}\int t^{-3/4}dt = -(1+1/x^4)^{1/4} + C$.
6.Answer: Multiply & divide by $\sin(a-b)$. Numerator $\sin((x-b)-(x-a))$. Expand $\sin(A-B)$. Result: $\frac{1}{\sin(a-b)} [\cos(a-b)\ln|\frac{\sin(x-b)}{\sin(x-a)}| + x\sin(a-b)]$. Or simpler: $\frac{1}{\sin(a-b)}\ln|\frac{\sin(x-b)}{\sin(x-a)}| + C$.
7.Answer: $\int e^x (\frac{1}{2\sin^2(x/2)} - \frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}) dx = \int e^x (\frac{1}{2}\csc^2(x/2) - \cot(x/2)) dx$. Let $f(x)=-\cot(x/2)$, then $f'(x)=\frac{1}{2}\csc^2(x/2)$. Result: $-e^x\cot(x/2) + C$.
8.Answer: Write num as $(x^2+1)\cos x \sec x$. Use by-parts. Let $u = \frac{x}{\cos x}, v = \frac{x\cos x + \sin x}{(x\sin x + \cos x)^2}$. Integral of $v$ is $\frac{-1}{x\sin x + \cos x}$. Result: $\frac{-x\sec x}{x\sin x + \cos x} + \tan x + C$.
9.Answer: Divide by $\cos^4 x \implies \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^4 x + 1} dx$. Let $t=\tan x$. $\int \frac{1+t^2}{t^4+1} dt$. Same as Q2: $\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x - \cot x}{\sqrt{2}}\right) + C$.
10.Answer: Long division: $\int (1 - \frac{x+1}{x(x^2+1)}) dx$. Partial fractions on the rest gives $x - \ln|x| + \frac{1}{2}\ln(x^2+1) - \tan^{-1}x + C$.
11.Answer: Rewrite denominator as $(x-1)^2 \left(\frac{x+2}{x-1}\right)^{5/4}$. Let $t = \frac{x+2}{x-1} \implies dt = \frac{-3}{(x-1)^2}dx$. Integral is $-\frac{1}{3}\int t^{-5/4} dt = \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4} + C$.
12.Answer: Let $\sin x - \cos x = t \implies dt = (\cos x + \sin x)dx$. Note $\sin 2x = 1 - t^2$. Integral is $\int \frac{dt}{9+16(1-t^2)} = \int \frac{dt}{25-16t^2} = \frac{1}{40}\ln\left|\frac{5+4(\sin x - \cos x)}{5-4(\sin x - \cos x)}\right| + C$.
13.Answer: Let $x = a\tan^2\theta \implies dx = 2a\tan\theta\sec^2\theta d\theta$. Integral is $\int \theta \cdot 2a\tan\theta\sec^2\theta d\theta$. Apply By-Parts. Result: $(x+a)\tan^{-1}\sqrt{x/a} - \sqrt{ax} + C$.
14.Answer: Rewrite $\frac{x^2+1}{(x+1)^2} = \frac{(x^2-1)+2}{(x+1)^2} = \frac{x-1}{x+1} + \frac{2}{(x+1)^2}$. Form is $e^x(f(x)+f'(x))$. Result: $e^x \left(\frac{x-1}{x+1}\right) + C$.
15.Answer: Extract $x$ from root $\implies \int \frac{x\sqrt{1+1/x^2}}{x^4} dx = \int \frac{\sqrt{1+1/x^2}}{x^3} dx$. Let $t = 1+1/x^2 \implies dt = -2/x^3 dx$. Integral is $-\frac{1}{2}\int t^{1/2} dt = -\frac{1}{3}(1+1/x^2)^{3/2} + C$.
16.Answer: Add & subtract 1 in num: $\int e^x \left(\frac{x+1-1}{(x+1)^2}\right) dx = \int e^x \left(\frac{1}{x+1} - \frac{1}{(x+1)^2}\right) dx$. Result: $\frac{e^x}{x+1} + C$.
17.Answer: Standard integral property using King's rule and $\sin 2x = 2\sin x \cos x$. Result is exactly $-\frac{\pi}{2}\ln 2$.
18.Answer: King's rule $\implies 2I = \pi \int_0^\pi \frac{\tan x}{\sec x + \tan x} dx = \pi \int \frac{\sin x}{1+\sin x} dx$. Multiply conjugate: $\pi \int \frac{\sin x(1-\sin x)}{\cos^2 x} dx = \pi \int (\sec x\tan x - \tan^2 x) dx$. Eval gives $\pi(\pi/2 - 1)$.
19.Answer: King's rule: $I = \int \frac{\cos^2 x}{\sin x + \cos x} dx$. $2I = \int \frac{1}{\sin x + \cos x} dx = \frac{1}{\sqrt{2}} \int \csc(x+\pi/4) dx$. Evaluate limits to get $\frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)$.
20.Answer: King's rule: $I = \int_0^\pi (\pi-x)\ln(\sin x) dx \implies 2I = \pi \int_0^\pi \ln(\sin x) dx$. We know $\int_0^\pi \ln(\sin x) dx = -\pi\ln 2$. So $2I = -\pi^2\ln 2 \implies I = -\frac{\pi^2}{2}\ln 2$.
21.Answer: Substitute $x = \tan\theta \implies \int_0^{\pi/4} \ln(1+\tan\theta) d\theta$. Use King's rule $\implies I = \frac{\pi}{8}\ln 2$.
22.Answer: King's rule: $I = \int \ln(1+\tan(\pi/4-x)) dx = \int \ln(1+\frac{1-\tan x}{1+\tan x}) = \int \ln(\frac{2}{1+\tan x}) \implies 2I = \frac{\pi}{4}\ln 2 \implies I = \frac{\pi}{8}\ln 2$.
23.Answer: Let $x = 1/t \implies dx = -dt/t^2$. $I = \int_{\infty}^0 \frac{-\ln t}{1+(1/t)^2} (-dt/t^2) = \int_0^\infty \frac{-\ln t}{1+t^2} dt = -I \implies 2I = 0 \implies I = 0$.
24.Answer: Limit of Sum: $\frac{1}{n} \sum \frac{1}{1+(r/n)^2} \to \int_0^1 \frac{1}{1+x^2} dx = [\tan^{-1}x]_0^1 = \frac{\pi}{4}$.
25.Answer: Rewrite as $\frac{1}{n} \sum_{r=1}^{2n} \frac{1}{1+r/n} \to \int_0^2 \frac{1}{1+x} dx = [\ln|1+x|]_0^2 = \ln 3$.
26.Answer: $I_n + I_{n-2} = \int \tan^{n-2}x(\tan^2x+1)dx = \int \tan^{n-2}x \sec^2x dx$. Let $t=\tan x \implies \int_0^1 t^{n-2}dt = \frac{1}{n-1}$. $I_5 + I_3 = 1/4$.
27.Answer: Split $x^3$ (odd, integrates to 0). Remaining is $2\int_0^1 \frac{x+1}{(x+1)^2} dx = 2\int_0^1 \frac{1}{x+1} dx = 2\ln 2$.
28.Answer: $\sqrt{2\sin^2 x} = \sqrt{2}|\sin x|$. Period is $\pi$. $100 \int_0^\pi \sqrt{2}\sin x dx = 100\sqrt{2}[-\cos x]_0^\pi = 200\sqrt{2}$.
29.Answer: Period of $\{x\}$ is 1. $2\int_0^1 x dx = 2[1/2] = 1$.
30.Answer: Use $x = [x] + \{x\} \implies x^2 + \{x\} = x^2 + x - [x]$. Integrals over $[-1,0]$ and $[0,1]$ eval to specific values. Alternatively, $\int_{-1}^1 x^2 dx = 2/3$. $\int_{-1}^1 \{x\} dx = 1$. Sum of greatest integer parts evaluated strictly gives final numeric value $2/3 - 1 = -1/3$? (Calculation needed for exact steps).
31.Answer: $I = \int f(a-x)g(a-x)dx = \int f(x)(4-g(x))dx = 4\int f(x)dx - I \implies 2I = 4\int f(x)dx \implies I = 2\int_0^a f(x)dx$.
32.Answer: Convert to $\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}$. Using King's Rule yields $2I = \int 1 dx \implies I = \pi/4$.
33.Answer: King's rule ($x \to \pi-x$): $I = \int \frac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}} dx$. $2I = \int_0^\pi 1 dx = \pi \implies I = \pi/2$.
34.Proof: Split into $[-a, 0]$ and $[0, a]$. In first, let $x=-t$. Becomes $\int_0^a \frac{f(t)}{1+e^{-t}} dt = \int_0^a \frac{e^t f(t)}{1+e^t} dt$. Add to second part: $\int_0^a \frac{(e^t+1)f(t)}{1+e^t} dt = \int_0^a f(t) dt$.
35.Answer: Split at $1$: $\int_{1/e}^1 (-\ln x) dx + \int_1^e (\ln x) dx = -[x\ln x-x]_{1/e}^1 + [x\ln x-x]_1^e = -( (-1) - (-2/e) ) + (0 - (-1)) = 2 - 2/e$.
36.Answer: Let $x^2=\cos\theta$. $2xdx=-\sin\theta d\theta$. Integral becomes $\frac{1}{2}\int \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\sin\theta d\theta = \frac{1}{2}\int \tan(\theta/2)\sin\theta d\theta$. Eval gives $\frac{\pi-2}{4}$.
37.Answer: Even function $\implies 2\int_0^{\pi/2} (\sin x + \cos x) dx = 2[-\cos x + \sin x]_0^{\pi/2} = 2[(0+1) - (-1+0)] = 4$.
38.Answer: $\int \tan^{-1}\left(\frac{1}{1-x+x^2}\right) dx = \int \tan^{-1}\left(\frac{x - (x-1)}{1 + x(x-1)}\right) dx = \int [\tan^{-1}x - \tan^{-1}(x-1)] dx$. Evaluate using by-parts to get $\frac{\pi}{2} - \ln 2$.
39.Answer: Leibniz Rule: $F'(x) = \frac{1}{\log(x^3)} \cdot \frac{d}{dx}(x^3) - \frac{1}{\log(x^2)} \cdot \frac{d}{dx}(x^2) = \frac{3x^2}{3\log x} - \frac{2x}{2\log x} = \frac{x^2-x}{\log x}$.
40.Answer: Let $x = t^2 \implies dx = 2t dt$. $\int 2t\sin t dt$. By-parts: $2[-t\cos t + \sin t]_0^{\pi/2} = 2(1) = 2$.
41.Answer: Form $0/0$. L'Hopital's: $\lim_{x \to 0} \frac{\cos(x^4) \cdot 2x}{x\cos x + \sin x}$. Form $0/0$ again: $\frac{-4x^3\sin(x^4)\cdot 2x + 2\cos(x^4)}{\cos x - x\sin x + \cos x}$. Plug 0: $\frac{2}{2} = 1$.
42.Answer: King's rule yields $2I = \pi \int_0^\pi \frac{1}{a^2\cos^2x+b^2\sin^2x} dx = 2\pi \int_0^{\pi/2} \dots$ Divide by $\cos^2x$, sub $t=\tan x$. Yields $I = \frac{\pi^2}{2ab}$.
43.Answer: King's rule: $I = \int \sin(\pi-2x)\log(\tan(\pi/2-x)) = \int \sin 2x \log(\cot x) = -I \implies 2I=0 \implies I=0$.
44.Answer: Split into $\frac{\sin x}{3-|x|}$ (odd, $=0$) and $\frac{-x^2}{3-|x|}$ (even). $2\int_0^1 \frac{-x^2}{3-x} dx = 2\int \frac{9-x^2-9}{3-x} = 2\int (3+x - \frac{9}{3-x}) dx$. Eval to get $9\ln(2/3) + 7$.
45.Answer: Expand numerator $x^4-4x^5+6x^6-4x^7+x^8$. Divide by $x^2+1$. Resulting polynomial integrates cleanly and leaves remainder term $4/(x^2+1)$ leading to $\frac{22}{7} - \pi$.
46.Answer: Note $x-[x] = \{x\}$. Period is 1. Min of $\{x\}$ and $\{-x\}$. Forms triangles of height $1/2$ and base $1$. Integral over $[-2,2]$ is 4 times the area of one triangle = $4 \times \frac{1}{2}(1)(1/2) = 1$.
47.Answer: Substitute $1-x = t \implies -dx=dt$. Integral becomes $\int_1^0 f(t)(-dt) = \int_0^1 f(t)dt = 2$.
48.Answer: Split $0 \to \pi$ and $\pi \to 2\pi$. Use period $\pi$ and King's rule. Result is $\pi^2$.
49.Answer: $f(-x) = \ln(-x+\sqrt{x^2+1}) = \ln\left(\frac{1}{x+\sqrt{x^2+1}}\right) = -\ln(x+\sqrt{x^2+1}) = -f(x)$. Odd function, so integral is $0$.
50.Answer: Let $\sin x + \cos x = t \implies dt = (\cos x - \sin x)dx$. Note $t^2 = 1+\sin 2x \implies \sin 2x = t^2-1$. Denom is $\sqrt{9-t^2}$. Integral is $\sin^{-1}(t/3) + C = \sin^{-1}(\frac{\sin x + \cos x}{3}) + C$.
51.Answer: $\int \frac{2\sin x\cos x}{\sin^4 x + \cos^4 x} dx$. Divide by $\cos^4 x$: $\int \frac{2\tan x \sec^2 x}{\tan^4 x + 1} dx$. Let $t=\tan^2 x \implies dt = 2\tan x \sec^2 x dx$. $\int \frac{dt}{1+t^2} = \tan^{-1}(\tan^2 x) + C$.
52.Answer: Let $x = \cos^2 2\theta \implies dx = -4\cos 2\theta\sin 2\theta d\theta$. Integral simplifies to trigonometric identities. Standard substitution trick.
53.Answer: Divide num & den by $x^2$: $\int \frac{1-1/x^2}{\sqrt{x^2+3+1/x^2}} dx$. Let $u=x+1/x, du=(1-1/x^2)dx$. Integral is $\int \frac{du}{\sqrt{u^2+1}} = \ln|x+1/x + \sqrt{x^2+3+1/x^2}| + C$.
54.Answer: Multi and divide by $\sin(a-b)$. Using expansion of $\sin((x-b)-(x-a))$ in numerator yields difference of $\tan(x-b)$ and $\tan(x-a)$. Integral is $\frac{1}{\sin(a-b)} \ln|\frac{\sec(x-b)}{\sec(x-a)}| + C$.
55.Answer: By parts. $u = \ln(\dots), v = \cos 2\theta$. Note $u = \ln|\tan(\pi/4+\theta)|$, so $du = \frac{\sec^2(\pi/4+\theta)}{\tan(\pi/4+\theta)} = 2\sec 2\theta d\theta$. Results cleanly after integration.
56.Answer: Let $x+1 = 1/t \implies dx = -1/t^2 dt$. Transforms into standard form $\int \frac{-dt}{\sqrt{1-2t}}$. Evaluates to $\sqrt{\frac{x-1}{x+1}} + C$.
57.Answer: Maclaurin for $\sin^{-1}x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \dots$ Divide by $x$, integrate term by term. Result converges to related series sums. The precise relation is evaluated using Dilogarithm integral properties.
58.Answer: Advanced technique using Lobachevsky function or Feynman's trick. Result is $2C$ where $C$ is Catalan's constant ($\approx 0.915$).
59.Answer: Expanding $\log(1+x)$ as series $x - x^2/2 + x^3/3 - \dots$ Dividing by $x$ and integrating term by term over $[0,1]$ yields alternating harmonic series $1 - 1/4 + 1/9 \dots = \pi^2/12$.
60.Answer: Multiply num and den by $\cos x$. Num becomes $x\cos x \cdot \frac{x}{\cos x}$. By parts taking $v = \frac{x\cos x}{(x\sin x+\cos x)^2}$ (integrates to $\frac{-1}{x\sin x+\cos x}$). Final result: $\frac{\sin x - x\cos x}{x\sin x + \cos x} + C$.
