8.∫ u′(x)dx
u(x) = ln|u(x)|+C9.∫
(αu(x) +β)nu′(x)dx=(αu(x) +β)n+1
α(n+ 1) +C10.∫ u′(x)v(x)−v′(x)u(x)
v^2 (x) dx=u(x)
v(x)+C∫ u′(x)v(x)−u(x)v′(x)
u(x)v(x) dx= ln|u(x)
v(x)|+C∫ u′(x)v(x)−u(x)v′(x)
u^2 (x) +v^2 (x) dx= tan− 1 u(x)
v(x)+C∫ u′(x)v(x)−u(x)v′(x)
u^2 (x)−v^2 (x) dx=1
2 ln|u(x)−v(x)
u(x) +v(x)|+C∫ u′(x)dx
√
u^2 (x) +α= ln|u(x) +√
u^2 (x) +α|+C∫ u(x)dx
(u(x) +α)(u(x) +β)=
α
α−β∫ dx
u(x) +α−β
α−β∫ dx∫ u(x) +β, α^6 =β
dx
u(x) +α−α∫ dx
(u(x) +α)^2 , β=α∫ u′(x)dx
αu^2 (x) +βu(x)=1
βln|u(x)
αu(x) +β|+C∫ u′(x)dx
u(x)√
u^2 (x)−α^2=α^1 sec−^1 u(αx)+C∫ u′(x)dx
α^2 +β^2 u^2 (x)=1
αβtan− 1 βu(x)
α +C∫ u′(x)dx
α^2 u^2 (x)−β^2 =1
2 αβln|αu(x)−β
αu(x) +β|+C∫
f(sinx)dx= 2∫
f( 2 u
1 +u^2) du
1 +u^2 , u= tanx
2∫
f(sinx)dx=∫
f(u)√du
1 −u^2, u= sinx∫
f(cosx)dx= 2∫
f( 1 −u 2
1 +u^2) du
1 +u^2 , u= tanx
2∫
f(cosx)dx=−∫
f(u)√ 1 du−u 2 , u= cosx∫
f(sinx,cosx)dx=∫
f(u,√
1 −u^2 )√du
1 −u^2, u= sinx∫
f(sinx,cosx)dx= 2∫
f(
2 u
1 +u^2 ,1 −u^2
1 +u^2)
du
1 +u^2 , u= tanx
2∫
f(x,√
α+βx)dx=^2 β∫
f(u (^2) −α
β , u
)
udu, u^2 =α+βx
27.
∫
f(x,
√
α^2 −x^2 )dx=α
∫
f(αsinu, acosu) cosu du, x=αsinu
Appendix C