题目

Problem

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

\int \frac{9x^5}{\sqrt{x^3+1}}\,dx=2(x^3+1)^k(x^3-A)+c

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

\int \frac{9x^5}{\sqrt{x^3+1}}\,dx=2(x^3+1)^k(x^3-A)+c

其中 $k$ 和 $A$ 是待求常数， $c$ 为任意常数。

解答

Let

u=\sqrt{x^3+1}

Then

u^2=x^3+1

Differentiate with respect to $x$ :

2u\frac{du}{dx}=3x^2

dx=\frac{2u}{3x^2}\,du

Now

\begin{aligned} \int \frac{9x^5}{\sqrt{x^3+1}}\,\mathrm{d}x &=\int \frac{9x^5}{u}\cdot \frac{2u}{3x^2}\,\mathrm{d}u \\ &=\int 6x^3\,\mathrm{d}u \end{aligned}

Since

u^2=x^3+1

we have

x^3=u^2-1

Therefore

\int \frac{9x^5}{\sqrt{x^3+1}}\,\mathrm{d}x =6\int (u^2-1)\,\mathrm{d}u

Integrate:

\begin{aligned} 6\int (u^2-1)\,\mathrm{d}u &=6\left(\frac13u^3-u\right)+c \\ &=2u^3-6u+c \end{aligned}

Substitute back $u=\sqrt{x^3+1}$ :

2u^3-6u+c =2(x^3+1)^{\frac32}-6(x^3+1)^{\frac12}+c

Factorise:

\begin{aligned} 2(x^3+1)^{\frac32}-6(x^3+1)^{\frac12} &=2(x^3+1)^{\frac12}\left((x^3+1)-3\right) \\ &=2(x^3+1)^{\frac12}(x^3-2) \end{aligned}

Hence

\int \frac{9x^5}{\sqrt{x^3+1}}\,\mathrm{d}x =2(x^3+1)^{\frac12}(x^3-2)+c

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