题目
Problem
Find the exact value of ∫ 0 6 x ( x + 1 ) x 2 + 4 d x \displaystyle\int_0^6\frac{x(x+1)}{x^2+4}\,\mathrm{d}x ∫ 0 6 x 2 + 4 x ( x + 1 ) d x
(6)
题目中文翻译
求 ∫ 0 6 x ( x + 1 ) x 2 + 4 d x \displaystyle\int_0^6\frac{x(x+1)}{x^2+4}\,\mathrm{d}x ∫ 0 6 x 2 + 4 x ( x + 1 ) d x
(6)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
解法一
思路
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先做多项式除法，把被积函数拆成常数、对数型与反正切型三项。
答题过程
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Rewrite the integrand:
x ( x + 1 ) x 2 + 4 = 1 + x − 4 x 2 + 4 = 1 + x x 2 + 4 − 4 x 2 + 4 . \begin{align*}
\frac{x(x+1)}{x^2+4}
=&\,1+\frac{x-4}{x^2+4}\\[4mm]
=&\,1+\frac{x}{x^2+4}-\frac4{x^2+4}.
\end{align*} x 2 + 4 x ( x + 1 ) = = 1 + x 2 + 4 x − 4 1 + x 2 + 4 x − x 2 + 4 4 . The derivative of x 2 + 4 x^2+4 x 2 + 4 2 x 2x 2 x d ( tan − 1 ( x / 2 ) ) / d x = 2 / ( x 2 + 4 ) \mathrm{d}(\tan^{-1}(x/2))/\mathrm{d}x=2/(x^2+4) d ( tan − 1 ( x /2 )) / d x = 2/ ( x 2 + 4 )
I = [ x + 1 2 ln ( x 2 + 4 ) − 2 tan − 1 x 2 ] 0 6 = 6 + 1 2 ln 40 − 2 tan − 1 3 − 1 2 ln 4 = 6 + 1 2 ln 10 − 2 tan − 1 3 . \begin{align*}
I
=&\,\bigg[x+\frac12\ln(x^2+4)
-2\tan^{-1}\frac x2\bigg]_0^6\\[4mm]
=&\,6+\frac12\ln40-2\tan^{-1}3-\frac12\ln4\\[4mm]
=&\,\boxed{6+\frac12\ln10-2\tan^{-1}3}.
\end{align*} I = = = [ x + 2 1 ln ( x 2 + 4 ) − 2 tan − 1 2 x ] 0 6 6 + 2 1 ln 40 − 2 tan − 1 3 − 2 1 ln 4 6 + 2 1 ln 10 − 2 tan − 1 3 .
解法二
思路
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也可作 x = 2 tan θ x=2\tan\theta x = 2 tan θ sec 2 θ \sec^2\theta sec 2 θ tan 2 θ = sec 2 θ − 1 \tan^2\theta=\sec^2\theta-1 tan 2 θ = sec 2 θ − 1
答题过程
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Put x = 2 tan θ x=2\tan\theta x = 2 tan θ
d x = 2 sec 2 θ d θ , x 2 + 4 = 4 sec 2 θ . \begin{align*}
\mathrm{d}x=&\,2\sec^2\theta\,\mathrm{d}\theta,\\[4mm]
x^2+4=&\,4\sec^2\theta.
\end{align*} d x = x 2 + 4 = 2 sec 2 θ d θ , 4 sec 2 θ . The limits x = 0 , 6 x=0,6 x = 0 , 6 θ = 0 , tan − 1 3 \theta=0,\tan^{-1}3 θ = 0 , tan − 1 3
I = ∫ 0 tan − 1 3 ( 2 tan 2 θ + tan θ ) d θ = ∫ 0 tan − 1 3 ( 2 sec 2 θ − 2 + tan θ ) d θ = [ 2 tan θ − 2 θ + ln ( sec θ ) ] 0 tan − 1 3 . \begin{align*}
I
=&\,\int_0^{\tan^{-1}3}
(2\tan^2\theta+\tan\theta)\,\mathrm{d}\theta\\[4mm]
=&\,\int_0^{\tan^{-1}3}
(2\sec^2\theta-2+\tan\theta)\,\mathrm{d}\theta\\[4mm]
=&\,\big[2\tan\theta-2\theta+\ln(\sec\theta)\big]_0^{\tan^{-1}3}.
\end{align*} I = = = ∫ 0 t a n − 1 3 ( 2 tan 2 θ + tan θ ) d θ ∫ 0 t a n − 1 3 ( 2 sec 2 θ − 2 + tan θ ) d θ [ 2 tan θ − 2 θ + ln ( sec θ ) ] 0 t a n − 1 3 . Here sec θ > 0 \sec\theta>0 sec θ > 0 sec θ = 1 + 3 2 = 10 \sec\theta=\sqrt{1+3^2}=\sqrt{10} sec θ = 1 + 3 2 = 10
I = 6 + 1 2 ln 10 − 2 tan − 1 3 . \boxed{I=6+\frac12\ln10-2\tan^{-1}3}. I = 6 + 2 1 ln 10 − 2 tan − 1 3 . 