⑩串紫学 Teaching Plan on Advanced Mathematics 例8求2dc(a>0) x+a 解:令x=tant→= asec tdt t∈ asec tdt x +al sect Isectdt=In(sect+ tant)+C 2 x√x+a In+ x+ll yTianjin Polytechnic University Teaching Plan on Advanced Mathematics ( 0). 1 2 2  +  dx a x a 令 x = atant dx a tdt 2  = sec = +  dx x a 2 2 1 a tdt a t 2 sec sec 1   =  sectdt = ln(sec t + tant) +C t a x 2 2 x + a ln . 2 2 C a x a a x +        + = +          − 2 , 2 t 例8 求 解：