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受欢迎的微积分 >

integral from 1 to 9 of 1/(xsqrt(x^2+9))

解答

\int_{1}^{9} \frac{1}{x x ^{2} + 9} d x

解答

\frac{1}{3} (ln (tan (\frac{1}{2} arctan (3))) - ln (tan (\frac{1}{2} arctan (\frac{1}{3}))))

+1

十进制

0.49699 \dots

求解步骤

\int_{1}^{9} \frac{1}{x x ^{2} + 9} d x

使用三角换元法

= \int_{a r c t a n (\frac{1}{3})}^{a r c t a n (3)} \frac{sec ( u )}{3 tan ( u )} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{a r c t a n (\frac{1}{3})}^{a r c t a n (3)} \frac{sec ( u )}{tan ( u )} d u

使用换元积分法

= \frac{1}{3} \cdot \int_{t a n (\frac{a r c t a n ( \frac{1}{3} )}{2})}^{t a n (\frac{a r c t a n ( 3 )}{2})} \frac{1}{v} d v

使用积分常见定则:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} [ln ∣ v ∣]_{t a n (\frac{a r c t a n ( \frac{1}{3} )}{2})}^{t a n (\frac{a r c t a n ( 3 )}{2})}

带入上下限计算后得:

ln (tan (\frac{1}{2} arctan (3))) - ln (tan (\frac{1}{2} arctan (\frac{1}{3})))

= \frac{1}{3} (ln (tan (\frac{1}{2} arctan (3))) - ln (tan (\frac{1}{2} arctan (\frac{1}{3}))))

作图

流行的例子

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t a y l or \frac{1}{1 + 2 x}

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x \to 3 lim ((x - 0) \cdot (x - 3))

泰勒 sin(x), pi/6

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\int (e^{y^{3} + 2}) d y

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\int_{0}^{4} (x) d x