∫√tanx+1dx

∫dx/(1+tanx)∫dx/(1+tanx)∫dx/(1+tanx)

求∫tanx/(1-(tanx)^2)dx求∫tanx/(1-(tanx)^2)dx求∫tanx/(1-(tanx)^2)dx=∫cosxsinx/[cos^2x-sin^2x]dx=∫sinx/[c

∫tanx(tanx＋1)dx∫tanx(tanx＋1)dx∫tanx(tanx＋1)dx∫tanx(tanx＋1)dx=∫（tan²x+tanx）dx=∫（sec²x-1+tan

∫dx/(1+tanX)=?∫dx/(1+tanX)=?∫dx/(1+tanX)=?令t=tanx原式=∫1/[(1+t)(1+t^2)]dt=(1/2)∫1/(1+t)dt-(1/2)∫(t-1)/

求不定积分∫[(√tanx)+1]/[(cosx)^2]dx求不定积分∫[(√tanx)+1]/[(cosx)^2]dx求不定积分∫[(√tanx)+1]/[(cosx)^2]dx∫[(√tanx)+

求不定积分∫{√[(tanx)+1]}/[(cosx)^2]dx求不定积分∫{√[(tanx)+1]}/[(cosx)^2]dx求不定积分∫{√[(tanx)+1]}/[(cosx)^2]dx注意到d

∫（1-tanx）/（1+tanx）dx求導?∫（1-tanx）/（1+tanx）dx求導?∫（1-tanx）/（1+tanx）dx求導?原式=∫(tanπ/4-tanx)/(1+tanπ/4tanx

积分∫1/(1+tanx)dx积分∫1/(1+tanx)dx积分∫1/(1+tanx)dx如图：

∫dx/1+tanx怎么求∫dx/1+tanx怎么求∫dx/1+tanx怎么求∫dx/1+tanx=∫（cosx/（sinx+cosx））dx=A令B=∫（sinx/（sinx+cosx））dx则A+

∫sec²x/1+tanxdx∫sec²x/1+tanxdx∫sec²x/1+tanxdx?

∫ln（1+tanx）dx=∫ln（1+tanx）dx=∫ln（1+tanx）dx=如果是求定积分的话就好了∫[0,π/4]ln(1+tanx)dx换元π/4-t=x=-∫[π/4,0]ln[1+(1

求不定积分∫((tanx)^4-1)dx,求不定积分∫((tanx)^4-1)dx,求不定积分∫((tanx)^4-1)dx,原式=∫(tan²x+1)(tan²x-1)dx=∫s

∫sin^2x(1+tanx)dx∫sin^2x(1+tanx)dx∫sin^2x(1+tanx)dxsinx^2(1+tanx)=(1-cosx^2)(1+tanx)=1+tanx-cosx^2-s

求不定积分?∫(tanx-1)^2dx求不定积分?∫(tanx-1)^2dx求不定积分?∫(tanx-1)^2dx∫(tanx-1)^2dx=∫(tan^2(x)+1-2tanx)dx=∫(sec^2

∫(secx/1+tanx)^2dx∫(secx/1+tanx)^2dx∫(secx/1+tanx)^2dx原式=∫(sec²x)/(1+tanx)²dx=∫1/(1+tanx)&

∫dx/(sinx+tanx)∫dx/(sinx+tanx)∫dx/(sinx+tanx)三角函数万能公式令tanx/2=t(secx/2)^2dx/2=dtdx=2dt/(t^2+1)∫dx/(si

∫(tanx+x)dx∫(tanx+x)dx∫(tanx+x)dx1.∫(tanx+x)dx＝∫tanxdx+∫xdx2.∫tanxdx,令u=cosx,du=－sinxdx．∫tanxdx=-ln|

∫(tanx)^4dx∫(tanx)^4dx∫(tanx)^4dx∫(tanx)^4dx=∫(tanx)^2[(tanx)^2+1-1]dx=∫(tanx)^2(secx)^2dx-∫(tanx)^2

∫(tanx)^2dx∫(tanx)^2dx∫(tanx)^2dx=∫(sec²x-1)dx=∫(sec²x)dx-∫dx=tanx-x+C∫x(tanx)^2dx＝∫x[(sec

1.∫[sin(2x)]^2dx2.∫(1-tanx)/(1+tanx)dx1.∫[sin(2x)]^2dx2.∫(1-tanx)/(1+tanx)dx1.∫[sin(2x)]^2dx2.∫(1-ta