题目

Problem

Use the substitution $u=\sqrt{x^3+1}$ to show that

\begin{align*} &\,\int \frac{9x^5}{\sqrt{x^3+1}}\,\mathrm{d}x\\ &\,=2(x^3+1)^k(x^3-A)+c \end{align*}

where $k$ and $A$ are constants to be found and $c$ is an arbitrary constant.

(5)

题目中文翻译

使用代换 $u=\sqrt{x^3+1}$ ，证明

\begin{align*} &\,\int \frac{9x^5}{\sqrt{x^3+1}}\,\mathrm{d}x\\ &\,=2(x^3+1)^k(x^3-A)+c \end{align*}

其中 $k$ 和 $A$ 是待求常数， $c$ 为任意常数。

(5)

解答

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按题目指定的平方根代换，将 $3x^2\,\mathrm{d}x$ 整体换成 $2u\,\mathrm{d}u$ 。积分后先提出 $2u$ ，再代回原变量，与给定形式比较。

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Use the required substitution

u=\sqrt{x^3+1}.

Then $u^2=x^3+1$ and $x^3=u^2-1$ . Differentiating,

2u\frac{\mathrm{d}u}{\mathrm{d}x}=3x^2,

3x^2\,\mathrm{d}x=2u\,\mathrm{d}u.

Denote the required integral by $I$ . Write $9x^5=3x^3(3x^2)$ and substitute:

\begin{align*} I=&\,\int\frac{3x^3}{u}\,(3x^2\,\mathrm{d}x)\\ =&\,\int\frac{3x^3}{u}\,(2u\,\mathrm{d}u)\\ =&\,6\int(u^2-1)\,\mathrm{d}u. \end{align*}

Integrating in $u$ ,

\begin{align*} I=&\,2u^3-6u+c\\ =&\,2u(u^2-3)+c. \end{align*}

Now reverse the substitution. Since $u=\sqrt{x^3+1}$ and $u^2-3=x^3-2$ ,

I=2(x^3+1)^{1/2}(x^3-2)+c.

This is the required form, with

\boxed{k=\frac12,\qquad A=2}.