Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times the inverse tangent of the quantity x squared divided by 4, plus C one fourth times the natural log of x to the 4th power plus 16, plus C one fourth times the inverse tangent of the quantity x squared divided by 4, plus C the product of negative one fourth and 1 over the quantity squared of x squared plus 16, plus C

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]

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