Indefinite Integrals
Definite Integrals
Solve Using Substitution
Integrals of Particular Functions
Tougher Integrals
100

∫ 2x dx

x2+C

100

o2 xdx

4

100

∫ 5sec2(5x+1) dx

tan(5x+1)+C

100

∫xn dx

(xn+1/n)+C

100

1/(x^2 - x + 1)

(3*sqrt(3))/8 * arctan(2/sqrt(3) * (x - 1/2)) + C

200

∫ 3x+2 dx

(3x2/2)+2x+C

200

15 2x4+3 dx

6308/5 or 1261.6

200

∫ esin(ex) dx

-cos(ex)+C

200

∫sin(x)dx

-cos(x)+C

200

∫ sec3(x) dx

(ln(|tan(x) + sec(x)|) + sec(x)tan(x))/2 + C

300

∫  17+3xdx

17x+(3x5/5)+C

300

01 3x+7 dx

17/2 or 8.5

300

01 -12x2(4x3-1)3 dx

-20

300

∫ex dx

ex+C

300

-ππ (1 + 2⌊(x/π)⌋)ex2 (sec(x) - cos(x)) dx

0

400

∫ 6x2+12x+3 dx

3x3+6x2+3x+C

400

69.5 3x3+3x dx

5218.172

400

∫ sin(ln(x))/x dx

-cos(ln(x))+C

400

∫ sec(x)tan(x) dx

sec(x)+C

400

∫ tan3(ln(x)) / x dx

1/2 tan2(ln|x|) - ln|cos(ln|x|)| + C

500

∫ 6x9+23x+4x3 dx

(3x10/5)+(23x2/2)+x4+C

500

012 x2+34 dx

984

500

∫ (ex)/(ex+1) dx

ln(ex+1)+C

500

∫ csc2(x) dx

-cot(x)+C

500

-11 (1/x) * sqrt((1+x)/(1-x)) * ln((2x2+2x+1)/(2x2-2x+1)) dx

4π arccot(sqrt(phi))

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