Antiderivatives
U-Substitution
Definite Integrals
Particular solutions
(Finding c)
Bonus
100

∫(5π‘₯4 +8π‘₯3 βˆ’9π‘₯2 βˆ’1)𝑑π‘₯

π‘₯5 + 2π‘₯4 βˆ’3π‘₯3 βˆ’ π‘₯ + 𝐢

100

∫Cos(5x+3)(5) dx 

Sin(5x+3)+C

100

∫-68 10 dx

140

100

dy/dx = x-4; y(2)=0

y = x2/2 - 4x + 6

100

∫-14(7ex) dx

7e7-7e-14

200

∫(π‘₯1/4 + 1/√π‘₯ +1/π‘₯ βˆ’ 1/π‘₯2)𝑑π‘₯

(4/5)x5/4 - 2√x + ln lxl + 1/π‘₯ + c

200

∫(4x6+7)12(24x5) dx

(4x6+7)13/13  +C

200

∫02 (4x3-3x2+2x) dx

12

200

𝑑𝑦/𝑑π‘₯ = 3/π‘₯2; 𝑦(1)=2

𝑦=βˆ’3/π‘₯ +5

200


∫ 2/(1βˆ’π‘₯2)1/2 𝑑π‘₯

2sinβˆ’1 π‘₯+𝐢

300

∫(𝑒π‘₯ +secπ‘₯tanπ‘₯βˆ’tanπ‘₯) 𝑑π‘₯

𝑒π‘₯ +secπ‘₯+ln|cosπ‘₯|+𝐢

300

∫(7x+4)dx

(1/7)(7x+4)10/10 + c

300

∫-2-1(6/x3) dx

-9/4

300

y'=x2-1; y(3)=7

y=(x3/3)-x+1

300

∫8 sin(5x) dx

(-8/5) cos(5x) + C

400

 βˆ«(3π‘₯ + 4/(1+π‘₯2))𝑑π‘₯


(1/(ln3))3π‘₯ +4tanβˆ’1 π‘₯+𝐢

400

∫1/(1+(4x)2) dx

1/4 tan-1(4x)+C

400

∫02 (x/(1+2x2)dx

2/9

400

𝑦′ =sinπ‘₯; 𝑦(0)=5

𝑦=βˆ’cosπ‘₯+6

400

𝑑𝑧/𝑑𝑑 =cosπ‘‘βˆ’2sin𝑑; 𝑧(0)=5

𝑧(𝑑)=sin𝑑+2cos𝑑+3

500

∫ 5π‘₯3/(1+π‘₯4) 𝑑π‘₯


(5/4)ln|1+π‘₯4|+𝐢

500

∫sec7x tan x dx

(sec x)7/7 + C

500

∫0πœ‹ (sec2(x/3)) dx

3√3

500

dy/dx=2/(x+1); y(0)=2

2ln|π‘₯+1|+2

500

d2y/ dx2= 4-3x ;y'(0)=5 & y(0)=7

y'=4x-((3x2)/2)+5

y''=2x2-(1/2)x3+5x+7

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