不定积分题目:如图以下是我的解法请指教
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不定积分题目:如图
![](http://img.wesiedu.com/upload/1/80/180e7d884ae0121c288e565533f94f0b.jpg)
以下是我的解法
![](http://img.wesiedu.com/upload/e/77/e77c8b86e4b02bb0b8791ef85edac88a.jpg)
请指教
![](http://img.wesiedu.com/upload/1/80/180e7d884ae0121c288e565533f94f0b.jpg)
以下是我的解法
![](http://img.wesiedu.com/upload/e/77/e77c8b86e4b02bb0b8791ef85edac88a.jpg)
请指教
![不定积分题目:如图以下是我的解法请指教](/uploads/image/z/4265477-53-7.jpg?t=%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E9%A2%98%E7%9B%AE%EF%BC%9A%E5%A6%82%E5%9B%BE%E4%BB%A5%E4%B8%8B%E6%98%AF%E6%88%91%E7%9A%84%E8%A7%A3%E6%B3%95%E8%AF%B7%E6%8C%87%E6%95%99)
I have a nice method that don't require you to set up equations
just do some simple substitutions
再问: Excuse me, could you explain more about the integral of
再问: Why d x/(x^2+9)=0?
再答: since the interval of the integral is symmetric about the y - axis together with the integrand x/(x² + 9), which is an odd function that means f(- x) = - f(x), it's rotational symmetric about the origin for the interval [- a,0], the value of the bounded area is negative(or positive) for the interval [0, a], the value of the bounded area is positive(or negative) as a result, the areas cancels each other and the total value of the integral is zero. ∫(- a→a) ƒ(x) dx = 0 if ƒ(x) is an odd function on the contrary, if the integrand is an even function, i.e. f(- x) = f(x) this function is symmetric about the y - axis which means the two areas bounded in the two intevals are totally identical therefore we have ∫(- a→a) ƒ(x) dx = 2∫(0→a) ƒ(x) dx i.e. 1/(x² + 9) is an even function
just do some simple substitutions
![](http://img.wesiedu.com/upload/e/28/e28fe36f3b61459da5529872c5cbee76.jpg)
再问: Excuse me, could you explain more about the integral of
再问: Why d x/(x^2+9)=0?
再答: since the interval of the integral is symmetric about the y - axis together with the integrand x/(x² + 9), which is an odd function that means f(- x) = - f(x), it's rotational symmetric about the origin for the interval [- a,0], the value of the bounded area is negative(or positive) for the interval [0, a], the value of the bounded area is positive(or negative) as a result, the areas cancels each other and the total value of the integral is zero. ∫(- a→a) ƒ(x) dx = 0 if ƒ(x) is an odd function on the contrary, if the integrand is an even function, i.e. f(- x) = f(x) this function is symmetric about the y - axis which means the two areas bounded in the two intevals are totally identical therefore we have ∫(- a→a) ƒ(x) dx = 2∫(0→a) ƒ(x) dx i.e. 1/(x² + 9) is an even function