Question

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 1

`\int\limits {(x)/(x^4 + 16)} \, dx = (1)/(8)*arctan((x^2)/(4)) + c`

Explanation:

Given

`\int\limits {(x)/(x^4 + 16)} \, dx`

Required

Solve

Let

`u = (x^2)/(4)`

Differentiate

`du = 2 * (x^(2-1))/(4)\ dx`

`du = 2 * (x)/(4)\ dx`

`du = (x)/(2)\ dx`

Make dx the subject

`dx = (2)/(x)\ du`

The given integral becomes:

`\int\limits {(x)/(x^4 + 16)} \, dx = \int\limits {(x)/(x^4 + 16)} \, * (2)/(x)\ du`

`\int\limits {(x)/(x^4 + 16)} \, dx = \int\limits {(1)/(x^4 + 16)} \, * (2)/(1)\ du`

`\int\limits {(x)/(x^4 + 16)} \, dx = \int\limits {(2)/(x^4 + 16)} \,\ du`

Recall that:
`u = (x^2)/(4)`

Make
`x^2` the subject

`x^2= 4u`

Square both sides

`x^4= (4u)^2`

`x^4= 16u^2`

Substitute
`16u^2` for
`x^4` in
`\int\limits {(x)/(x^4 + 16)} \, dx = \int\limits {(2)/(x^4 + 16)} \,\ du`

`\int\limits {(x)/(x^4 + 16)} \, dx = \int\limits {(2)/(16u^2 + 16)} \,\ du`

Simplify

`\int\limits {(x)/(x^4 + 16)} \, dx = \int\limits {(2)/(16)* (1)/(8u^2 + 8)} \,\ du`

`\int\limits {(x)/(x^4 + 16)} \, dx = (2)/(16)\int\limits {(1)/(u^2 + 1)} \,\ du`

`\int\limits {(x)/(x^4 + 16)} \, dx = (1)/(8)\int\limits {(1)/(u^2 + 1)} \,\ du`

In standard integration

`\int\limits {(1)/(u^2 + 1)} \,\ du = arctan(u)`

So, the expression becomes:

`\int\limits {(x)/(x^4 + 16)} \, dx = (1)/(8)\int\limits {(1)/(u^2 + 1)} \,\ du`

`\int\limits {(x)/(x^4 + 16)} \, dx = (1)/(8)*arctan(u)`

Recall that:
`u = (x^2)/(4)`

`\int\limits {(x)/(x^4 + 16)} \, dx = (1)/(8)*arctan((x^2)/(4)) + c`