4.2.14.2.2对于第二类的不定积分，有a(c+)+b-ac+b(+)2+q-)k(c2+pc+q)kt10(t2 + d2) + b1(t2+d2)k这里b=b-,d=Vq-,=a+号t1-当k=1时In(t2 + d°) +C,dt2t2+d21t1dt+Carctan-ddt2+d2t11当k>1时，dt(t2 + d2)k-i + C,(t2 + d2)k2(1 - k)1dt可用递推公式求出+d2)kt2返回全屏关闭退出6/16
4.2.1 4.2.2 éu1aؽȩ, k ax + b (x2 + px + q) k = a(x + p 2 ) + b − ap 2 (x + p 2 ) 2 + q − p2 4 k = a t (t 2 + d2) k + b1 1 (t 2 + d2) k , ùp b1 = b − ap 2 , d = q q − p2 4 , t = x + p 2 . k = 1 , Z t t 2 + d2 dt = 1 2 ln(t 2 + d 2 ) + C, Z 1 t 2 + d2 dt = 1 d · arctan t d + C k > 1 , Z t (t 2 + d2) k dt = 1 2(1 − k) · 1 (t 2 + d2) k−1 + C, Z 1 (t 2 + d2) k dt ^4íúª¦Ñ. 6/16 kJ Ik J I £ ¶ '4 òÑ
4.2.14.2.2例2求dr.a3+1-+2本=十解根据前面的例子可知++11da=ln+1|+C..&+1-t+-+212dtdr(t =a22-+1V3t22+31+Carctan212In(2-+1)+V3arctann2于是V3(α+1)212+Cdlnrarctan一2r363-a+1+1返回全屏关闭退出中7/16
4.2.1 4.2.2 ~ 2 ¦ Z 1 x3 + 1 dx. ) âc¡~f 1 x3+1 = 1 3 · 1 x+1 + 1 3 · −x+2 x2−x+1. Z 1 x + 1 dx = ln |x + 1| + C. Z −x + 2 x2 − x + 1 dx = Z −t + 3 2 t 2 + ( √ 3 2 ) 2 dt (t = x − 1 2 ) = − 1 2 ln t 2 + ( √ 3 2 ) 2 ! + 3 2 · 2 √ 3 arctan 2 √ 3 t + C = − 1 2 ln(x 2 − x + 1) + √ 3 arctan 2 √ 3 (x − 1 2 ) + C u´ Z 1 x3 + 1 dx = 1 6 ln (x + 1)2 x2 − x + 1 + √ 3 3 arctan 2 √ 3 (x − 1 2 ) + C. 7/16 kJ Ik J I £ ¶ '4 òÑ