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∫tAn√(1+x^2)xDx/√(1+x^2)

方法不唯一,仅供参考

答案在图片里

凑微分即可

=∫arctanx/x√(1-x²)darcsinx=∫√(1-x²)arctanx/x√(1-x²)dx=∫arctanx/x dx=∫arctanx/x dx2 =2∫arctanxdx=2(xarctanx-∫xdarctanx)=2(xarctanx-∫x/(1+x2)dx)=2(xarctanx-∫x/(1+x2)d(x2+1)) =2(xarctanx-∫2x.x/(1+x2)d(x2+1)...

lim(x->∞) (1/x^2)arctan{ (x^2+x+1)/[(x+1)(x-2)] } =lim(x->∞) (1/x^2) .lim(x->∞) arctan{ (x^2+x+1)/[(x+1)(x-2)] } =lim(x->∞) (1/x^2) .lim(x->∞) arctan{ (1+1/x+1/x^2)/[(1+1/x)(1-2/x)] } =lim(x->∞) (1/x^2) . (π/4) =0

∫(arctanx)/(1+x^2) dx =∫(arctanx) d(arctanx) =(1/2)(arctanx)²+C

函数是初等函数, 在x=0与x=±1处没有定义, 所以,仅有x=0与x=±1这三个间断点。 f(x)=1/x²·arctan[x/(x²-1)] lim(x→0)f(x) =lim(x→0)1/x²·x/(x²-1) =lim(x→0)1/[x·(x²-1)] =∞ ∴x=0是第二类无穷间断点。 lim(x→1-)f(x) ...

=∫x/(1+x^2)dx+∫e^(arctanx)/(1+x^2)dx =1/2·ln(1+x^2)+∫e^(arctanx)d(arctanx) =1/2·ln(1+x^2)+∫e^(arctanx)+C

(1+x^2)^0.5的不定积分怎么求解? ∫√(1+x^2)dx 令x=tant,则dx=d(tant)=sec^tdt t=arctanx 原式=∫√(1+tan^2t)*sec^2tdt =∫sec^3tdt =∫sect*sec^2tdt =∫sectd(tant) =sect*tant-∫tantd(sect) =sect*tant-∫tant*tant*sectdt =sect*tant-∫tan^2t*se...

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