搜搜考试网求一道不定积分的题……
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求一道不定积分的题……
看似不是很难.可是就是做不来=.=
∫ln(x^2+2)dx
不要跳步.
看似不是很难.可是就是做不来=.=
∫ln(x^2+2)dx
不要跳步.
原式=xln(x²+2)-∫xdln(x²+2)
=xln(x²+2)-∫2x²/(x²+2)dx
=xln(x²+2)-2∫[1-2/(x²+2)]dx
=xln(x²+2)-2x+4∫dx/(x²+2)
令x=√2a
dx=√2da
则∫dx/(x²+2)
=√2∫da/(2a²+2)
=√2/2∫da/(a²+1)
=√2/2*arctan(a²+1)
=√2/2*arctan(x²/2+1)
所以原式=xln(x²+2)-2x+2√2arctan(x²/2+1)+C
=xln(x²+2)-∫2x²/(x²+2)dx
=xln(x²+2)-2∫[1-2/(x²+2)]dx
=xln(x²+2)-2x+4∫dx/(x²+2)
令x=√2a
dx=√2da
则∫dx/(x²+2)
=√2∫da/(2a²+2)
=√2/2∫da/(a²+1)
=√2/2*arctan(a²+1)
=√2/2*arctan(x²/2+1)
所以原式=xln(x²+2)-2x+2√2arctan(x²/2+1)+C