Question

高数不定积分求解 最好有解题过程 可以写在纸上照下来 照片可以发到我的邮箱里yaoyao19930918@sina.com

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Answer 1

1
∫f'(x^3)dx=x^4-x+1
f'(x^3)=(x^4-x+1)'=4x^3-1
f'(x)=4x-1
f(x)=∫f'(x)dx=2x^2-x+c
2
d∫(f(2x+3t)dt /dx =(u=3t)
=(2/3)d∫f(2x+u)du/d(2x)=(2/3)f(2x+u)=(2/3)f(2x+3t)
3
f'(cosx)=sinx f'(x)=√(1-x^2)
f(x)=∫√(1-x^2)dx cosu=x dx=-sinudu =∫-sinu^2du=∫(cos2u-1)/2=sin2u/4-u/2+C
=x√(1-x^2)/2-arccosx/2+C
f(cosx)=sinxcosx/2-x/2+C

1
∫(2^x+3^x)^2dx=∫(4^x+6^x+9^x)dx=(1/ln4)4^x+(1/ln6)6^x+(1/ln9)9^x+C
2
∫dx/(sinxcosx)^2=∫4dx/sin2x=2∫dcos2x/[(1-cos2x)(1+cos2x)]=ln|(1+cos2x)/(1-cos2x)|+C
3
∫(1-1/x^2)√(x+1/x)dx=(2/3)√(x+1/x)^3+C
4
∫sin√xdx/√x=-cos(√x)+C
5
∫(1-1/x^2)e^(x+1/x)dx=e^(x+1/x)+C

6
∫dx/[cosx^2√(1-tanx^2)]=-∫dtanx/√1-tanx^2=-arcsin(tanx)+C

7
∫cotxdx/lnsinx= -∫dlnsinx/lnsinx=-ln|lnsinx|+C
8
∫x^2dx/[(x^3-1)^(2/3)=(x^3-1)^(1/3)+C

9
∫x^3dx/[1+(1-x^4)^(1/3)]=(1/4)∫d(x^4-1)/[1+(1-x^4)^(1/3)] 1-x^4=t
=(-1/4)∫dt/(1+t^(1/3)=(-3/4)∫t^(2/3)dt^(1/3)/(1+t^(1/3))
=(-1/4)∫[t^(1/3)-1]dt^(1/3) +(3/4)∫dt^(1/3)/(1+t^(1/3)
=(-1/8)t^(2/3)+(1/4)t^(1/3)+(3/4)ln|1+t^(1/3)|+C
=(-1/8)(1-x^4)^(2/3)+(1/4)(1-x^4)^(1/3)+(3/4)ln|1+(1-x^4)^(1/3)+C
10
∫dx/[√x+x^(1/4)]=∫4x^(3/4)/[x^1/4(x^(1/4)+1)]
=∫4x^(3/4)dx^(1/4)/x^1/4) -∫4^(3/4)dx/(1+x^(1/4))
=4∫x^(2/4)dx^(1/4) -4∫[1-x^(1/4)+x^(2/4)]dx^(1/4) +∫dx^(1/4)/(1+x^(1/4))
=(4/3)x^(3/4)-4x^(1/4)+2x^(1/2)-(4/3)x^(3/4)+ln(1=x^(1/4)+C

11
∫xarctanxdx/(1+x^2)

12
∫ln(x+√(1+x^2)dx=xln(x+√(1+x^2)) -∫xdln(x+√(1+x^2)) =xln(x+√(1+x^2))-√(x^2+1)+C
[ln(x+√(1+x^2)]'=[1+x/√(1+x^2)]/[x+√(1+x^2)]=1/√(1+x^2)

13
∫x^3e^x^2dx=(1/2)∫x^2de^(x^2)=(1/2)x^2e^(x^2)-(1/2)e^x^2+C
14
∫dx/[x√(x^2-1)]=∫xdx/√(x^2-1)-∫dx/x=√(x^2-1)-lnx+C

15
∫dx/√(x^2+1)^3 (tanu=x cosu=1/√(1+x^2),sinu=x/√(1+x^2))
=∫cosudu=sinu+C=x/√(1+x^2)+C