Q. 37 - 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.) 

$\int x^4{(x^3+1)}^2\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int x^4{(x^3+1)}^2\mathrm{d}x=\frac{1}{11}{x}^{11}+\frac{1}{8}{x}^8+\frac{1}{5}{x}^5$.

Step 1. Given Information

Solving the given integrals. 

$\int x^4{(x^3+1)}^2\mathrm{d}x$

Step 2. Solving the given integral using algebra.

$$\begin{gathered} \int x^4{(x^3+1)}^2\mathrm{d}x=\int x^4\{{\left(x^3\right)}^2+2·x^3·1+{\left(1\right)}^2\}\mathrm{d}x \\ \int x^4{(x^3+1)}^2\mathrm{d}x=\int x^4(x^6+2x^3+1)\mathrm{d}x \\ \int x^4{(x^3+1)}^2\mathrm{d}x=\int (x^4·x^6+2·x^4·x^3+x^4·1)\mathrm{d}x \\ \int x^4{(x^3+1)}^2\mathrm{d}x=\int (x^{10}+2x^7+x^4)\mathrm{d}x \end{gathered}$$

Step 3. After simplifying

$$\begin{gathered} \int x^4{(x^3+1)}^2\mathrm{d}x=\int x^{10}\mathrm{d}x+2\int x^7\mathrm{d}x+\int x^4\mathrm{d}x \\ \int x^4{(x^3+1)}^2\mathrm{d}x=\left[\frac{x^{10+1}}{10+1}\right]+\left[\frac{x^7}{7+1}\right]+\left[\frac{x^{4+1}}{4+1}\right] \\ \int x^4{(x^3+1)}^2\mathrm{d}x=\frac{x^{11}}{11}+\frac{x^8}{8}+\frac{x^5}{5} \\ \int x^4{(x^3+1)}^2\mathrm{d}x=\frac{1}{11}{x}^{11}+\frac{1}{8}{x}^8+\frac{1}{5}{x}^5 \end{gathered}$$