1 doc = x²+ A log | x+ √x²+A | + c (A = 0 ) 次の不定積分を計算せよ。 (1) | √ 1* 1-x dxc (2) 1 doc 2x+x+ | (以下フリースペース) り (2) 222+2x+1 = 2 (242) + 1 = 2 {(2+1)² - £} +1 =2(x+2)² + √√2x²+2x+1 = √2(2 + 2)²+= = √ √4(x++)² + | t=x+1/2 2+² de=dx T 5√224+2+1 db = √√√√4(4) de = H = = 1 圧 =2t dx = 2 12 4+2+1 1 du= 2x dt dt = du > √2) √42+1 de = √ √ dv √ log | V+ √√²++ | + C √√// log 12++ √4±² + 1 1+ J12+1 √/2log | 2x + 1 + √4(2+2) 41 | + C