∫dx/(x^2+9)^2
令x=3tan(u),则dx=3sec^2(u)du，u=arctan(x/3)
上式=∫3sec^2(u)du/（9tan^2(u)+9)^2
=∫3sec^2(u)du/81sec^4(u)
=1/27∫du/sec^2(u)=1/27∫cos^2(u)du=1/27∫1/2cos(2u)+1/2du
=1/27*[1/2*u+1/4sin(2u)]+C
=1/54*[u+sin(u)cos(u)]+C
=1/54*[arctan(x/3)+3x/(x^2+9)+C
点个采纳？
......不会
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