Q. 52 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int x^2\sqrt{x+1}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int x^2\sqrt{x+1}\mathrm{d}x=\frac{2}{7}{(\mathrm{x}+1)}^{7/2}-\frac{4}{5}{(\mathrm{x}+1)}^{5/2}+\frac{2}{3}{(\mathrm{x}+1)}^{3/2}+C$.

Step 1. Given Information

Solving the given integrals.

$\int x^2\sqrt{x+1}\mathrm{d}x$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=x+1 \\ \frac{du}{dx}=1 \\ du=dx \end{gathered}$$

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

$x=u-1$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int x^2\sqrt{x+1}\mathrm{d}x=\int {(u-1)}^2\sqrt{u}\mathrm{du} \\ \int {\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\int {\{{\left(\mathrm{u}\right)}^2-2\times \mathrm{u}\times 1+(1)}^2\}{\mathrm{u}}^{1/2}\mathrm{du} \\ \int {\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\int {(\mathrm{u}}^2-2\mathrm{u}+1){\mathrm{u}}^{1/2}\mathrm{du} \\ \int {\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\int {(\mathrm{u}}^2·{\mathrm{u}}^{1/2}-2\mathrm{u}·{\mathrm{u}}^{1/2}+1·{\mathrm{u}}^{1/2})\mathrm{du} \\ \int {\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\int {(\mathrm{u}}^{2+1/2}-2{\mathrm{u}}^{1+1/2}+{\mathrm{u}}^{1/2})\mathrm{du} \\ \int {\mathrm{x}}^2\sqrt{\mathrm{x}+1}\mathrm{dx}=\int {(\mathrm{u}}^{5/2}-2{\mathrm{u}}^{3/2}+{\mathrm{u}}^{1/2})\mathrm{du} \end{gathered}$$

Step 4. After simplification.

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