1.Answer: Let $u = x^2+5x-7$. $du = (2x+5)dx$. Integral is $\int \frac{du}{u} = \ln|x^2+5x-7| + C$.
2.Answer: Let $u = \tan^{-1}x$. $du = \frac{1}{1+x^2}dx$. Integral is $\int e^u du = e^u + C = e^{\tan^{-1}x} + C$.
3.Answer: Let $u = \log x$. $du = \frac{1}{x}dx$. Integral is $\int \sin u du = -\cos u + C = -\cos(\log x) + C$.
4.Answer: Let $u = \sqrt{x}$. $du = \frac{1}{2\sqrt{x}}dx \implies 2du = \frac{dx}{\sqrt{x}}$. Integral is $\int 2\sec^2 u du = 2\tan u + C = 2\tan(\sqrt{x}) + C$.
5.Answer: $\int \frac{1}{x(1+\log x)} dx$. Let $u = 1+\log x$. $du = \frac{1}{x}dx$. Integral is $\int \frac{du}{u} = \ln|1+\log x| + C$.
6.Answer: Divide by $e^x$: $\int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$. Let $u = e^x + e^{-x}$. $du = (e^x - e^{-x})dx$. Integral is $\ln(e^x + e^{-x}) + C$.
7.Answer: Let $u = \sin x$. $du = \cos x dx$. Integral is $\int \frac{du}{1+u^2} = \tan^{-1}u + C = \tan^{-1}(\sin x) + C$.
8.Answer: Let $u = 1+\sqrt{x}$. $du = \frac{1}{2\sqrt{x}}dx \implies 2du = \frac{dx}{\sqrt{x}}$. Integral is $\int \frac{2du}{u} = 2\ln|1+\sqrt{x}| + C$.
9.Answer: Let $u = x^3$. $du = 3x^2 dx \implies \frac{du}{3} = x^2 dx$. Integral is $\frac{1}{3}\int \frac{du}{1+u^2} = \frac{1}{3}\tan^{-1}(x^3) + C$.
10.Answer: Let $u = a^2\cos^2 x + b^2\sin^2 x$. $du = (b^2-a^2)\sin 2x dx$. Integral is $\frac{1}{b^2-a^2} \ln|a^2\cos^2 x + b^2\sin^2 x| + C$.
11.Answer: Let $u = x^{10} + 10^x$. $du = (10x^9 + 10^x \ln 10) dx$. Integral is $\int \frac{du}{u} = \ln|x^{10} + 10^x| + C$.
12.Answer: Let $u = 1+\log x$. $du = \frac{1}{x} dx$. Integral is $\int \sec^2 u du = \tan u + C = \tan(1+\log x) + C$.
13.Answer: Let $u = x^4$. $du = 4x^3 dx$. Integral is $\frac{1}{4} \int \frac{du}{\sqrt{1-u^2}} = \frac{1}{4}\sin^{-1}(x^4) + C$.
14.Answer: Let $u = x e^x$. $du = e^x(1+x) dx$. Integral is $\int \frac{du}{\cos^2 u} = \int \sec^2 u du = \tan u + C = \tan(x e^x) + C$.
15.Answer: Multiply by $e^x/e^x$: $\int \frac{e^x}{e^{2x}+1} dx$. Let $u = e^x$. $du = e^x dx$. Integral is $\int \frac{du}{u^2+1} = \tan^{-1}(e^x) + C$.
16.Answer: $\int \sin x(1-\cos^2 x)\cos^2 x dx$. Let $u=\cos x, du=-\sin x dx$. $\int -(1-u^2)u^2 du = \int (u^4-u^2) du = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$.
17.Answer: $\int (\frac{1+\cos 2x}{2})^2 dx = \frac{1}{4} \int (1+2\cos 2x + \cos^2 2x) dx = \frac{1}{4} \int (1+2\cos 2x + \frac{1+\cos 4x}{2}) dx = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$.
18.Answer: Let $x-a=t$. $\int \frac{\sin(t+a)}{\sin t} dt = \int (\cos a + \sin a \cot t) dt = t\cos a + \sin a \ln|\sin t| + C = (x-a)\cos a + \sin a \ln|\sin(x-a)| + C$.
19.Answer: $\int \frac{1}{2\cos^2(x/2)} dx = \frac{1}{2} \int \sec^2(x/2) dx = \frac{1}{2} \frac{\tan(x/2)}{1/2} + C = \tan(x/2) + C$.
20.Answer: $\frac{1}{2} \int (\cos 4x - \cos 12x) dx = \frac{\sin 4x}{8} - \frac{\sin 12x}{24} + C$.
21.Answer: $\sqrt{\sin^2 x + \cos^2 x + 2\sin x \cos x} = \sqrt{(\sin x + \cos x)^2} = \sin x + \cos x$. Integral is $-\cos x + \sin x + C$.
22.Answer: $\int \frac{\cos^2 x - \sin^2 x}{\cos x + \sin x} dx = \int (\cos x - \sin x) dx = \sin x + \cos x + C$.
23.Answer: $\int \tan^2 x (\sec^2 x - 1) dx = \int \tan^2 x \sec^2 x dx - \int (\sec^2 x - 1) dx = \frac{\tan^3 x}{3} - \tan x + x + C$.
24.Answer: $\int \sec^2 x (1+\tan^2 x) dx$. Let $u=\tan x \implies \int (1+u^2) du = \tan x + \frac{\tan^3 x}{3} + C$.
25.Answer: $\int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx = \int (\sec^2 x + \csc^2 x) dx = \tan x - \cot x + C$.
26.Answer: $\int \frac{\cos^2 x - \sin^2 x}{(\sin x + \cos x)^2} dx = \int \frac{\cos x - \sin x}{\cos x + \sin x} dx$. Let $u=\cos x + \sin x$, integral is $\ln|\sin x + \cos x| + C$.
27.Answer: Let $t = x+a \implies x = t-a$. $\int \frac{\sin(t-2a)}{\sin t} dt = \int (\cos 2a - \sin 2a \cot t) dt = x\cos 2a - \sin 2a \ln|\sin(x+a)| + C$.
28.Answer: $\int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} dx = \int (\tan x \sec^2 x + \frac{1}{\sin x \cos x}) dx = \frac{\tan^2 x}{2} + \ln|\tan x| + C$.
29.Answer: $\frac{1}{2} \int \cos 2x (\cos 4x + \cos 2x) dx = \frac{1}{4} \int (\cos 6x + \cos 2x + 1 + \cos 4x) dx = \frac{\sin 6x}{24} + \frac{\sin 2x}{8} + \frac{x}{4} + \frac{\sin 4x}{16} + C$.
30.Answer: $\sqrt{1+\sin 2x} = \sin x + \cos x$. Integral is $\int \frac{\sin x - \cos x}{\sin x + \cos x} dx = -\ln|\sin x + \cos x| + C$.
31.Answer: $\int \frac{dx}{(x+2)^2 + 9} = \frac{1}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C$.
32.Answer: $\int \frac{dx}{\sqrt{9 - (x+2)^2}} = \sin^{-1}\left(\frac{x+2}{3}\right) + C$.
33.Answer: $\frac{1}{2}\int \frac{2x-5+9}{x^2-5x+6} dx = \frac{1}{2}\ln|x^2-5x+6| + \frac{9}{2}\int \frac{dx}{(x-2.5)^2 - 0.25}$. Final: $\frac{1}{2}\ln|x^2-5x+6| + \frac{9}{2}\ln\left|\frac{x-3}{x-2}\right| + C$.
34.Answer: $\int \frac{2x+4-3}{\sqrt{x^2+4x+5}} dx = 2\sqrt{x^2+4x+5} - 3\ln|x+2 + \sqrt{x^2+4x+5}| + C$.
35.Answer: Multiply num/den by $x^3$: $\int \frac{x^3}{x^4(x^4+1)} dx$. Let $u=x^4$. $\frac{1}{4}\int \frac{du}{u(u+1)} = \frac{1}{4}\ln\left|\frac{x^4}{x^4+1}\right| + C$.
36.Answer: $\int (1 - \frac{1}{x^2+1}) dx = x - \tan^{-1}x + C$.
37.Answer: Partial fractions: $\frac{A}{x-1} + \frac{Bx+C}{x^2+1}$. Gives $A=1/2, B=-1/2, C=1/2$. Integral is $\frac{1}{2}\ln|x-1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\tan^{-1}x + C$.
38.Answer: Partial fractions. $\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$. Gives $A=11/4, B=-5/2, C=-11/4$. Integral is $\frac{11}{4}\ln|x+1| + \frac{5}{2(x+1)} - \frac{11}{4}\ln|x+3| + C$.
39.Answer: Put $x^2 = y$ just for partial fraction split: $\frac{y}{(y+1)(y+4)} = \frac{-1/3}{y+1} + \frac{4/3}{y+4}$. Integral is $-\frac{1}{3}\tan^{-1}x + \frac{2}{3}\tan^{-1}(\frac{x}{2}) + C$.
40.Answer: $\int \sqrt{(x+1)^2+4} dx = \frac{x+1}{2}\sqrt{x^2+2x+5} + 2\ln|x+1+\sqrt{x^2+2x+5}| + C$.
41.Answer: Divide by $x^2$: $\int \frac{1+1/x^2}{x^2+1/x^2} dx = \int \frac{1+1/x^2}{(x-1/x)^2+2} dx$. Let $u=x-1/x$. Integral is $\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x-1/x}{\sqrt{2}}\right) + C$.
42.Answer: Let $u = \log x$. $\int \frac{du}{u^2-5u+6} = \int (\frac{1}{u-3} - \frac{1}{u-2}) du = \ln\left|\frac{\log x - 3}{\log x - 2}\right| + C$.
43.Answer: Let $u=e^x, dx=du/u$. $\int \frac{du}{u(u-1)(u-2)}$. Partial fractions yield $\frac{1}{2}\ln|e^x| - \ln|e^x-1| + \frac{1}{2}\ln|e^x-2| + C$.
44.Answer: Let $x+2=t^2 \implies dx=2t dt$. Integral is $\int \frac{2t dt}{(t^2-3)t} = 2\int \frac{dt}{t^2-3} = \frac{1}{\sqrt{3}}\ln\left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right| + C$.
45.Answer: Let $u = \sin x$. $\int \frac{du}{(1-u)(2-u)} = \int (\frac{1}{1-u} - \frac{1}{2-u}) du = -\ln|1-\sin x| + \ln|2-\sin x| + C$.
46.Answer: $u = \tan^{-1}x, v=x$. $\frac{x^2}{2}\tan^{-1}x - \frac{1}{2}\int \frac{x^2}{1+x^2} dx = \frac{x^2}{2}\tan^{-1}x - \frac{1}{2}(x - \tan^{-1}x) + C$.
47.Answer: $x^2(-\cos x) - \int 2x(-\cos x) dx = -x^2\cos x + 2(x\sin x - \int \sin x dx) = -x^2\cos x + 2x\sin x + 2\cos x + C$.
48.Answer: $I = \int \sec x \sec^2 x dx = \sec x \tan x - \int \sec x \tan^2 x dx = \sec x \tan x - \int \sec x(\sec^2 x - 1) dx$. $2I = \sec x \tan x + \ln|\sec x+\tan x|$. $I = \frac{1}{2}[\sec x \tan x + \ln|\sec x+\tan x|] + C$.
49.Answer: $\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx$. Here $f(x) = \frac{1}{x}$. Result: $\frac{e^x}{x} + C$.
50.Answer: $I = e^{2x}\frac{-\cos 3x}{3} - \int 2e^{2x} \frac{-\cos 3x}{3} dx$. Apply parts again to get $I = \frac{e^{2x}}{13}(2\sin 3x - 3\cos 3x) + C$.
51.Answer: Take $v=1$. $x\log(x^2+1) - \int x \frac{2x}{x^2+1} dx = x\log(x^2+1) - 2(x - \tan^{-1}x) + C$.
52.Answer: $\sin^{-1}(\dots) = 2\tan^{-1}x$. So $\int 2\tan^{-1}x dx = 2(x\tan^{-1}x - \frac{1}{2}\ln(1+x^2)) + C$.
53.Answer: $\int e^x \left(\frac{x+1-1}{(x+1)^2}\right) dx = \int e^x \left(\frac{1}{x+1} - \frac{1}{(x+1)^2}\right) dx = \frac{e^x}{x+1} + C$.
54.Answer: Let $x = \cos^2\theta$. $dx = -2\sin\theta\cos\theta d\theta$. Integral is $-\int \theta \sin 2\theta d\theta$. Use by-parts. Back substitute $x$. Result: $x\cos^{-1}\sqrt{x} - \frac{1}{2}\sqrt{x-x^2} + C$.
55.Answer: Let $\sin^{-1}x=t, x=\sin t, dx=\cos t dt$. $\int \frac{\sin t \cdot t}{\cos t} \cos t dt = \int t \sin t dt = -t\cos t + \sin t = x - \sqrt{1-x^2}\sin^{-1}x + C$.
56.Answer: Let $\log x = t \implies x=e^t, dx=e^t dt$. $\int e^t(\log t + \frac{1}{t^2}) dt = \int e^t(\log t - \frac{1}{t} + \frac{1}{t} + \frac{1}{t^2})dt = e^t(\log t - 1/t) = x(\log(\log x) - \frac{1}{\log x}) + C$.
57.Answer: $\int e^x \left(\frac{1+2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}\right) dx = \int e^x \left(\frac{1}{2}\sec^2(x/2) + \tan(x/2)\right) dx = e^x \tan(x/2) + C$.
58.Answer: Let $x^2=t, 2xdx=dt$. $\frac{1}{2}\int t e^t dt = \frac{1}{2} e^t(t-1) = \frac{1}{2} e^{x^2}(x^2-1) + C$.
59.Answer: $\int e^x (\sec x + \sec x \tan x) dx$. Here $f(x) = \sec x$. Result is $e^x \sec x + C$.
60.Answer: Let $\log x = t \implies x=e^t$. $\int e^t \sin t dt = \frac{e^t}{2}(\sin t - \cos t) = \frac{x}{2}(\sin(\log x) - \cos(\log x)) + C$.
61.Answer: Using King's rule ($x \to \pi/2-x$), $I = \int \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} dx$. Adding yields $2I = \int 1 dx = \pi/2 \implies I = \pi/4$.
62.Answer: King's rule: $I = \int_0^1 (1-x)(1-(1-x))^n dx = \int_0^1 (1-x)x^n dx = \int (x^n - x^{n+1}) dx = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}$.
63.Answer: King's rule: $I = \int_0^{\pi/2} \log(\cot x) dx$. $2I = \int \log(\tan x \cot x) dx = \int \log 1 dx = 0 \implies I = 0$.
64.Answer: $\sin^7 x$ is an odd function. Integral from $-a$ to $a$ is $0$.
65.Answer: $I = \int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x} dx \implies 2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. Let $u=\cos x \implies 2I = \pi [-\tan^{-1}(\cos x)]_0^\pi = \pi(\pi/4 - (-\pi/4)) = \pi^2/2 \implies I = \pi^2/4$.
66.Answer: $|x|$ is even. $2\int_0^1 x dx = 2[1/2] = 1$.
67.Answer: Roots are $1, -3$. Split at $1$. $\int_0^1 (3-2x-x^2) dx + \int_1^2 (x^2+2x-3) dx = 5/3 + 7/3 = 12/3 = 4$.
68.Answer: King's rule yields $2I = \int_0^{\pi/2} 1 dx \implies I = \pi/4$. Valid for any $n$.
69.Answer: $I = \int_0^\pi \log(1-\cos x) dx$. $2I = \int_0^\pi \log(\sin^2 x) dx = 2\int_0^\pi \log(\sin x) dx = 4\int_0^{\pi/2} \log(\sin x) dx = 4(-\frac{\pi}{2}\ln 2) = -2\pi\ln 2 \implies I = -\pi\ln 2$.
70.Answer: $f(-x) = (-x)^3 \sin^2(-x) = -x^3 \sin^2 x = -f(x)$. Odd function. Integral is $0$.
71.Answer: King's rule: $I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx$. $2I = \int_0^a 1 dx = a \implies I = a/2$.
72.Answer: $\int_0^1 0 dx + \int_1^{1.5} 1 dx = [x]_1^{1.5} = 1.5 - 1 = 0.5$.
73.Answer: $\int_0^\pi \frac{1+\cos 2x}{2} dx = [\frac{x}{2} + \frac{\sin 2x}{4}]_0^\pi = \pi/2$.
74.Answer: $x^3 \cos x + x$ is odd (integrates to 0). Remaining $\int_{-2}^2 1 dx = 4$.
75.Answer: King's rule ($x \to \pi/2-x$): $I = \int \sin(\pi-2x) \log(\cot x) dx = \int \sin 2x (-\log(\tan x)) dx = -I \implies 2I = 0 \implies I = 0$.
