Answer:
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c[/tex]
Step-by-step explanation:
Given
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx[/tex]
Required
Solve
Let
[tex]u = \frac{x^2}{4}[/tex]
Differentiate
[tex]du = 2 * \frac{x^{2-1}}{4}\ dx[/tex]
[tex]du = 2 * \frac{x}{4}\ dx[/tex]
[tex]du = \frac{x}{2}\ dx[/tex]
Make dx the subject
[tex]dx = \frac{2}{x}\ du[/tex]
The given integral becomes:
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du[/tex]
Recall that: [tex]u = \frac{x^2}{4}[/tex]
Make [tex]x^2[/tex] the subject
[tex]x^2= 4u[/tex]
Square both sides
[tex]x^4= (4u)^2[/tex]
[tex]x^4= 16u^2[/tex]
Substitute [tex]16u^2[/tex] for [tex]x^4[/tex] in [tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du[/tex]
Simplify
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du[/tex]
In standard integration
[tex]\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)[/tex]
So, the expression becomes:
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)[/tex]
Recall that: [tex]u = \frac{x^2}{4}[/tex]
[tex]\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c[/tex]
