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1/1+tAn^2x的不定积分

1/1+tan^2x的不定积分 =∫1/sec^2xdx =∫cos^2xdx =∫(1+co2x)/2dx =1/2 x+1/4sin2x+c

如图

∫ (tan2x + sec2x)² dx = ∫ (tan²2x + 2sec2xtan2x + sec²2x) dx = (1/2)∫ (sec²2x - 1 + 2sec2xta2x + sec²2x) d(2x) = (1/2)(2tan2x - 2x + 2sec2x) + C = tan2x + sec2x - x + C

∫tan²x/(1-sin²x) dx =∫tan²x/cos²x dx =∫tan²x*sec²x dx =∫tan²x d(tanx) =(1/3)tan³x + C

∫ sin^2x/(1+sin^2x)dx =∫ 1-1/(1+sin^2x)dx = x-∫ [1/cos^2x]/(1/cos^2x+tan^2x)dx = x-∫ [sec^2x]/(sec^2x + tan^2x)dx = x-∫ 1/(1 + 2tan^2x)dtanx =x- 1/√2 *∫ 1/(1 + (√2tanx)^2)d(√2tanx) = x-1/√2 * arctan(√2tanx) + C

凑微分法: 以上,请采纳。

(8)∫ ln(x²+1)dx = x ln(x²+1) - ∫ x * 2x/( x²+1) dx = x ln(x²+1) - 2 ∫ [ 1 - 1/( x²+1) ] dx = x ln(x²+1) - 2x + 2 arctanx + C (11)令√x=t,dx=dt²=2tdt ∫sin√xdx =2∫tsintdt =-2∫tdcost =-...

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