91影视

Solving the given integrals.

$鈭�x^2\sqrt{x+1}\mathrm{d}x$

Let

In this question in which integration by substitution works even though the integrand is not at all in the form localid="1648978706557" $f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)$. A clever change of variables will allow us to rewrite the integral so that it can be algebraically simplified.

$x=u-1$

$$\begin{gathered} 鈭�x^2\sqrt{x+1}\mathrm{d}x=鈭�{(u-1)}^2\sqrt{u}\mathrm{du} \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=鈭�{\{{\left(\mathrm{u}\right)}^2-2脳\mathrm{u}脳1+(1)}^2\}{\mathrm{u}}^{1/2}\mathrm{du} \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=鈭�{(\mathrm{u}}^2-2\mathrm{u}+1){\mathrm{u}}^{1/2}\mathrm{du} \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=鈭�{(\mathrm{u}}^2路{\mathrm{u}}^{1/2}-2\mathrm{u}路{\mathrm{u}}^{1/2}+1路{\mathrm{u}}^{1/2})\mathrm{du} \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=鈭�{(\mathrm{u}}^{2+1/2}-2{\mathrm{u}}^{1+1/2}+{\mathrm{u}}^{1/2})\mathrm{du} \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=鈭�{(\mathrm{u}}^{5/2}-2{\mathrm{u}}^{3/2}+{\mathrm{u}}^{1/2})\mathrm{du} \end{gathered}$$

$$\begin{gathered} 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=鈭�{\mathrm{u}}^{5/2}\mathrm{du}-2鈭�{\mathrm{u}}^{3/2}\mathrm{du}+鈭�{\mathrm{u}}^{1/2}\mathrm{du} \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\left[\frac{{\mathrm{u}}^{5/2+1}}{5/2+1}\right]-2\left[\frac{{\mathrm{u}}^{3/2+1}}{3/2+1}\right]+\left[\frac{{\mathrm{u}}^{1/2+1}}{1/2+1}\right]+C \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\left[\frac{{\mathrm{u}}^{7/2}}{7/2}\right]-2\left[\frac{{\mathrm{u}}^{5/2}}{5/2}\right]+\left[\frac{{\mathrm{u}}^{3/2}}{3/2}\right]+C \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\frac{2}{7}{\mathrm{u}}^{7/2}-2路\frac{2}{5}路{\mathrm{u}}^{5/2}+\frac{2}{3}{\mathrm{u}}^{3/2}+C \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\frac{2}{7}{\mathrm{u}}^{7/2}-\frac{4}{5}路{\mathrm{u}}^{5/2}+\frac{2}{3}{\mathrm{u}}^{3/2}+C \\ 鈭�{\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\frac{2}{7}{(\mathrm{x}+1)}^{7/2}-\frac{4}{5}{(\mathrm{x}+1)}^{5/2}+\frac{2}{3}{(\mathrm{x}+1)}^{3/2}+C \end{gathered}$$