Question

高等数学，不定积分

Answer 1

(1)令√x=t,则x=t²,dx=2tdt
原式=2∫tsintdt
=-2∫td(cost)
=-2(tcost-∫costdt)
=-2(tcost-sint)+C
=-2√xcos√x-2sin√x+C
(2)原式=1/2*∫arcsinxd(x²)
=1/2*[x²arcsinx-∫x²d(arcsinx)]
=1/2*[x²arcsinx-∫x²dx/√(1-x²)]~~~~①
令x=sint,t∈(-π/2,π/2),则t=arcsinx,√(1-x²)=cost,dx=costdt
∫x²dx/√(1-x²)
=∫sin²t*costdt/cost
=∫sin²tdt
=∫(1-cos2t)/2*dt
=1/2*∫dt-1/4*∫cos2td(2t)
=t/2-sintcost/2+C
=-x√(1-x²)/2+arcsinx/2+C
代入①得原式=1/4*[2x²arcsinx+x√(1-x²)-arcsinx]+C

Answer 2

【解析】
设F'(u)=f(u)，则
∫f(2x+3t)dt
=1/3·∫f(2x+3t)d(2x+3t)
=1/3·F(2x+3t)+C

∴d/dx[∫f(2x+3t)dt]
=d/dx[1/3·F(2x+3t)+C]
=1/3·d/dx[F(2x+3t)]
=1/3·f(2x+3t)·2
=2/3·f(2x+3t)

Answer 3

=5ln|x|+C