integral from 3 to 6 (x^4+4)/(x) dx=

To solve the given integral ∫(x^4 + 4)/x dx from 3 to 6, you can use the technique of integration known as the Fundamental Theorem of Calculus. However, before using this theorem, let's simplify the integrand.

Breaking down the fraction into two separate terms, we have:

∫(x^4 + 4)/x dx = ∫x^4/x dx + ∫4/x dx

Now, let's solve each term separately:

First term: ∫x^4/x dx
To simplify this term, we can divide x^4 by x, resulting in:

∫x^(4-1) dx = ∫x^3 dx

Integrating x^3 gives us:

(1/4) x^4

Second term: ∫4/x dx
The integral of 4/x can be evaluated directly as:

4 ∫(1/x) dx

The integral of (1/x) is the natural logarithm of |x|:

4 ln(|x|)

Now we can combine the two terms and evaluate the definite integral from 3 to 6:

∫(x^4 + 4)/x dx = ∫x^3 dx + ∫4/x dx
= (1/4) x^4 + 4 ln(|x|) [from 3 to 6]

Substituting the upper and lower limits:

= [(1/4) (6^4) + 4 ln(|6|)] - [(1/4) (3^4) + 4 ln(|3|)]

Simplifying further:

= [(1/4) (1296) + 4 ln(6)] - [(1/4) (81) + 4 ln(3)]

= (324 + 4 ln(6)) - (20.25 + 4 ln(3))

Therefore, the value of the integral from 3 to 6 is:

(324 + 4 ln(6)) - (20.25 + 4 ln(3)) ≈ 302.881