Q. 56 - 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.)

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

Solving the given integrals.

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

Step 2. Using the substitution method.

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

$x=u-1$

Step 3. This substitution changes the integral into

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

Step 4. After simlifying.

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