Antiderivatives

∫x

½ x²

∫ k

kx+C

∫x_ⁿ when n does not =-1

xⁿ⁺¹/n+1

∫1/x or x^-1

ln|x|+C

∫ e^x

e^x+C

∫e^kx

1/k e^kx +C

∫ a^x

a^x/ln(a) +C

∫sin(x)

-cos(x)+C

∫cos(x)

sin(x)+C

∫sec²(x)

tan(x)+C

∫sec(x)tan(x)

sec(x)+C

∫1/1+x²

arctan(x)+C

∫1/√1-x²

arcsin(x)+C

tan

sin/cos

derivative of ln(x+C)

1/X+C

∫ 1/x+C

ln(X+C)

ln(e)

1

ln(1)

0

ln(e^x)

X

U sub for indefinite integrals

interior function is u, du is in terms of dx and can have constants to be able to completely rewrite the integral in terms of u, take antiderivative then sub x back in

U sub for definite integrals

define u and du, adjust bounds to be u(a) and u(b), evaluate antiderivative in the bounds in terms of u, no need to sub back!

integration by parts formula

∫udv=uv-∫vdu

∫1dx

x+C

d/dx (∫_a^xf(t))

f(x)

d/dx (∫_x^af(t))

-f(x)

derivative of tan

sec²(x)

derivative of sec

sec(x)tan(x)

trig identity for constant squared plus variable squared

tan² (x) +1=sec² (x)

trig identity for constant squared - variable squared

1-sin² (x)=cos² (x)

what is tan² (x)=

sec² -1

what is sec equal to

1/cos

Steps for completing the square

add (b/2)² to your x²+Cx binomial, subtract (b/2)² from your C term, rewrite as a sqaured binomial plus the C-(b/2)² term

∫trig(ax)

1/a trig(ax)