fsinx^5dx-搜答案-大师作文网

∫x√(x-5)dx

令t=√(x-5)去求解x=5+t^2dx=2tdt原积分=∫(5+t^2)*t*2tdt=∫(10t^2+2t^4)dt=10/3*t^3+2/5*t^5+c将t=√(x-5)代回结果即可得到结果.

∫5^(3x)dx=1/3*∫5^(3x)d(3x)=1/3*5^3x/ln5+C=5^3x/(3ln5)+C

∫(0,1)5^x(dx)=

请稍等.再答：刚才写错了再答：再问：5^x的反倒数是1/（ln5）*5^x吗？再答：是呀(a^x)的导数是a^xlna

dy/dx-y=xy^5

1,令³√(3-5x)=t,则x=(3-t³)/5,那么dx=-3t²/5dx∫³√(3-5x)dx=∫t(-3t²/5)dt=-3/5∫t³

∫sinxcosx/cos^5dx=∫cosx/cosx^5dcosx=∫1/cosx^4dcosx=∫cosx^-4dcosx=-1/3cosx^-3+C

积分(sinx)平方(cosx)5次方dx=积分(sinx)平方(cosx)4次方dsinx=积分(sinx)平方(1-(sinx)平方)平方dsinx=积分(sinx)平方(1-2(sinx)平方+

∫x/（x^2+5）dx

∫x/（x^2+5）dx=1/2(ln|x^2+5|)+C

很简单啊d（y^5）/dy的分母跟dy/dx的分子约分后不就等于d（y^5）/dx吗再问：他们说是复合函数求导，怎么看呢？怎么理解啊？哪个是f(u),哪个是u呢？再答：y是函数g(y)=y^5的自变量

1+sinx=1+cos(π/2-x)=2cos²(π/4-x/2)1/(x²+4x-5)=1/[(x+5)(x-1)]=[1/(x-1)-1/(x+5)]·1/6(3x+1)/(

∫sin (4-5x) dx

[cos(4-5x)]/5+C

d(∫(sin^5 7x)dx)

若F(x)=∫f(x)dx则dF(x)=f(x)dx所以d(∫[sin(7x)]^5dx)={[sin(7x)]^5}dx

∫dx/x(x^5+4)

∫1/[x(x^5+4)]dx=¼∫[(x^5+4)-x^5]/[x(x^5+4)]dx=¼∫[1/x-x^4/(x^5+4)]dx=¼[∫1/xdx-1/5∫1/(x^

1、∫xsin(x^2)cos3(x^2)dx＝（1/2)∫sin(x^2)cos3(x^2)dx^2=(1/4)∫[sin4(x^2)-sin2(x^2)]dx^2=(1/4)[∫sin4(x^2)

∫x(sinx)∧5 dx

∫1/(3-5x)^2dxt=3-5xx=(3-t)/5∫1/(3-5x)^2dx=∫1/t^2d(3-t)/5=-1/5∫1/t^2dt=1/5*1/t+C=1/[5(3-5x)]+C

∫5^x*dx怎么做

(5^x)'=5^x*ln5所以原式=∫(1/ln5)*5^x*ln5dx=5^x/ln5+C

∫f(3x+5)dx=(1/3)×∫f(3x+5)d(3x)=(1/3)×∫f(3x+5)d(3x+5)=(1/3)F(3x+5)+C