问一个不定积分问题，题目如图？ 求问一道高数不定积分题，题目如下图所示
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\u8fd9\u4e00\u7c7b\u9898\u4e00\u822c\u7528\u4e07\u80fd\u516c\u5f0f\u5316\u89e3\u3002
\u5df2\u77e5\uff1a
2tan\u03b1
sin2\u03b1\uff1d
\u2014\u2014
1
(tan\u03b1)^2
1-(tan\u03b1)^2
cos2\u03b1\uff1d
\u2014\u2014
1
(tan\u03b1)^2;
\u4e0d\u59a8\u8bbe\uff1ax=2*arctg(y)
\u8fd9\u6837
2*y^2
dx\uff1d
\u2014\u2014
1
y^2;
1
\u2014\u2014
=(1
y^2)/2
1
(cosx)^2;
\u8fd9\u6837\u5f97\u5230\u79ef\u5206\u5f0f\u5b50\u222by^2dy
\u63a5\u4e0b\u6765\u4f1a\u7b97\u4e86\u5427\uff0c\u8bb0\u4f4f\u8fd9\u4e2a\u65b9\u6cd5\uff0c\u5bf9\u4e8e\u4e09\u89d2\u51fd\u6570\u79ef\u5206\u5f88\u6709\u7528
∫xf'(x) dx = ln(1+x^2) +c
两边求导
xf'(x) = 2x/(1+x^2)
f'(x) = 2/(1+x^2)
f(x) =∫ 2/(1+x^2) dx
= 2arctanx + C'
f(1)=2
2=2(π/4) +C'
C'= 2-π/2
f(x) = 2arctanx +2 -π/2
求解过程与结果如图所示
绛旓細鈭玿f'(x) dx = ln(1+x^2) +c 涓よ竟姹傚 xf'(x) = 2x/(1+x^2)f'(x) = 2/(1+x^2)f(x) =鈭 2/(1+x^2) dx = 2arctanx + C'f(1)=2 2=2(蟺/4) +C'C'= 2-蟺/2 f(x) = 2arctanx +2 -蟺/2 ...
绛旓細绛旀B 绗簩姝 浠=x³鏂规硶濡涓嬪浘鎵绀猴紝璇蜂綔鍙傝冿紝绁濆涔犳剦蹇細
绛旓細I = 鈭垰(x^2-a^2)dx = x鈭(x^2-a^2) - 鈭玿^2dx/鈭(x^2-a^2)= x鈭(x^2-a^2) - 鈭(x^2-a^2+a^2)dx/鈭(x^2-a^2)= x鈭(x^2-a^2) - I + a^2鈭玠x/鈭(x^2-a^2)2I = x鈭(x^2-a^2) + a^2鈭玠x/鈭(x^2-a^2)浠 x = secu锛 鍒 鈭玠x/...
绛旓細瀹氱Н鍒涓嶅畾绉垎鏁板棰橈紝鍝綅鏈嬪弸鑳藉憡璇夋垜涓嬭В绛旀楠ゅ晩锛岄鐩鍥 閫夋嫨棰橈紝绗竴棰樼Н鍒杩斿洖鐨勬槸涓涓甯告暟锛屽嵆杞寲涓篸C/dx=0锛屽父鏁版枩鐜囦负0銆 绗簩棰橈紝绉垎寰梐³/3=9锛屽嵆a=3銆 绗笁棰樼Н鍒嗗緱k脳2²/2=2锛宬=1銆 绗洓棰樺悓绗竴棰樸 绗簲棰橈紝鍥犱负(lnx)'=1/x銆 绗叚棰...
绛旓細濡備笅
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绛旓細杩囩▼閲囩敤鍒嗛儴绉垎娉 绛旀鍦ㄥ浘鐗囦笂锛岀偣鍑诲彲鏀惧ぇ銆傛弧鎰忚鐐归噰绾筹紝璋㈣阿
绛旓細= 6鈭 [v⁴ + v³ + v² + v + 1/(v - 1) + 1] dv = 6[(1/5)v⁵ + (1/4)v⁴ + (1/3)v³ + (1/2)v² + ln|v - 1| + v| + C = (6/5)x^(5/6) + (3/2)x^(2/3) + 2鈭歺 + 3x^(1/3) + 6^(1/6...
绛旓細绠鍗曡绠椾竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細瑙ｅ涓嬪浘鎵绀
