x/(ax+b)dx的不定积分正推导怎么推?x^2/(ax+b)dx呢?
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x/(ax+b)dx的不定积分正推导怎么推?x^2/(ax+b)dx呢?
x/(ax+b)dx的不定积分正推导怎么推?x^2/(ax+b)dx呢?
x/(ax+b)dx的不定积分正推导怎么推?x^2/(ax+b)dx呢?
x/(ax+b)dx=1/a^2*ax/(ax+b)dax=1/a^2(1-b/(ax+b))dax=1/a^2*d(ax+b)-b/a^2*d(ax+b)/(ax+b)
积分得(ax+b)/a^2-b/a^2*ln(ax+b)
x^2/(ax+b)dx化成1/a^3*((ax+b)^2-b^2-2abx)/(ax+b)dax ,以下计算跟第一中情况基本一样
∫x/(ax+b) dx
= (1/a)*∫[(ax+b)-b]/(ax+b) dx
= (1/a)*∫ dx - (b/a)*∫1/(ax+b) dx
= (1/a)*x - (b/a)(1/a)*∫1/(ax+b)*d(ax+b)
= x/a - (b/a²)ln|ax+b| + C
∫x²/(ax+b) dx
= (1/a)...
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∫x/(ax+b) dx
= (1/a)*∫[(ax+b)-b]/(ax+b) dx
= (1/a)*∫ dx - (b/a)*∫1/(ax+b) dx
= (1/a)*x - (b/a)(1/a)*∫1/(ax+b)*d(ax+b)
= x/a - (b/a²)ln|ax+b| + C
∫x²/(ax+b) dx
= (1/a)*∫x[(ax+b)-b]/(ax+b) dx
= (1/a)*∫x dx - (b/a)*∫x/(ax+b) dx
= (1/a)*x²/2 - (b/a)(1/a)*∫[(ax+b)-b]/(ax+b) dx
= x²/(2a) - (b/a²)*∫ dx + (b²/a²)(1/a)*∫1/(ax+b) d(ax+b)
= x²/(2a) - bx/a² + (b²/a³)ln|ax+b| + C
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其实我搞不懂常数变易的推导，也没学过，只是我以前高中的时候了解过一点点∫dx/(ax^2+b)^n=x/[2b(n-1)(ax^2+b)^(n-1)]+(2n-3)/[