求不定积分详细过程(求不定积分，求详细过程)

求不定积分，求详细过程

The L.C.M. of 1/3 and 1/2 is 1/6

So let x = t^6 then dx = (6t^5)dt and we have

∫ x^(1/3) / [x(√x + x^(1/3))] dx

= ∫ t² / [t^6(t³+t²)] * (6t^5)dt

= 6∫ t/(t³+t²) dt

= 6∫ [(1+t)-t] / [t(1+t)] dt

= 6∫ 1/t dt - 6∫ 1/(1+t) dt

= 6ln|t| - 6ln|1+t| + C

= 6ln|x^6| - 6ln|1+x^(1/6)| + C

= ln|x| - 6ln|1+x^(1/6)| + C

Let t = √(2x+3) => t² = 2x+3 => 2tdt = 2dx => tdt = dx

∫ [2-√(2x+3)]/(1-2x) dx

= ∫ (2-t)/{1-2[(t²-3)/2]} tdt

= ∫ (2-t)/(4-t²) tdt

= ∫ (2-t)/[(2-t)(2+t)] tdt

= ∫ t/(2+t) dt

= ∫ [(2+t)-2]/(2+t) dt

= ∫ dt - 2∫ dt/(2+t)

= t - 2ln|2+t| + C

= √(2x+3) - 2ln| 2+√(2x+3) | + C

∫u/（sinu）²du=∫ud（-cotu）=-ucotu+∫cotudu=-ucotu+ln|sinu|+C

用分部积分法

①∫arcsinx dx

= xarcsinx - ∫x darcsinx

= xarcsinx - ∫x/√(1 - x²) dx

= xarcsinx + ∫1 /[2√(1 - x²)] d(1 - x²)

= xarcsinx + √(1 - x²) + C

② ∫ arccosx dx

=xarccosx-∫x darccosx

=xarccosx+∫x/√（1-x²）dx

=xarccosx-√（1-x²）+C

积分化为(省略积分号)=-(cosxdcosx)/(1+cos^2x)，如果你学积分时间不长，可做个代换：u=cosx更清楚

=-udu/(1+u^2)=-(1/2)d(1+u^2)/(1+u^2)=-(1/2)ln(1+u^2)+C，再将u换回cosx，在积分熟练后，这个代换就不必做，而直接计算