Issues With Simple Integration Problem

[swashbuckler77]

My Calc II class is currently doing Calc I integration review before we get started with integration by parts. I am using an online system that gives immediate feedback on my work and it has said that my answer is wrong. Not sure where I'm going wrong.

Homework Statement

Evaluate: $$\int \frac{(t+9)^2}{t^3} dx$$

Homework Equations

Sum rule: $$\int (f + g) dx = \int (f) dx + \int (g) dx$$
Common denominator rule: $$\frac{(a+b)}{c} = \frac{a}{c} + \frac{b}{c}$$

The Attempt at a Solution

1. Multiply out the numerator
$$(t+9)^2$$
$$t^2 + 18t + 81$$
2. Result
$$= \int \frac{t^2 + 18t + 81}{t^3} dx$$
3. Split up into three fractions by the common denominator rule provided
$$= \int ( \frac{t^2}{t^3} + \frac{18t}{t^3} + \frac{81}{t^3} ) dx$$
4. Simplify
$$= \int ( \frac{1}{t} + \frac{18}{t^2} + \frac{81}{t^3} ) dx$$
5. By the Sum Rule, I can rewrite the integral as follows
$$= \int \frac{1}{t} dx + \int \frac{18}{t^2} dx + \int \frac{81}{t^3} dx$$
6. Integrate each part
$$= ln(t) - \frac{18}{t} - \frac{81}{2t^2} + c$$
7. My final answer
$$\int \frac{(t+9)^2}{t^3} dx = ln(t) - \frac{18}{t} - \frac{81}{2t^2} + c$$

 

[Differentiate1]

Your answer is essentially correct.
You can use WolframAlpha to check it if you're unsure of any answer you have in the future.
http://www.wolframalpha.com/input/?i=integrate+(t+9)^2/t^3 (Your answer is simplified).

The mistake I see from your work is that you're integrating with respect to x. Your colonel (what you want to integrate), is in terms of t. Make sure they're consistent otherwise the system you're using to check answers won't understand what you want.

If you're curious as to what you typed will give, it will give this: {[(t+9)^2]/(t^3)}(x) + C

Source:

Calculus 3 student

 

[swashbuckler77]

Haha, thanks for catching that, Differentiate1- I should have seen that while writing my post! I had also checked with Wolfram but wanted to verify with a human just in case. Looks like I need to email the site administrator. Thank you.

 

[ehild]

Your result excludes x<0 as ln(x) is defined only for x>0. The integral of 1/x is ∫1/x dx =ln|x|.

ehild

 

[Mark44]

Your colonel (what you want to integrate), is in terms of t.

Colonel? That's a military rank, just below general. Did you mean "kernel"? In English the two words are pronounced the same, but I am not familiar with "kernel" being used in integration. A couple of words that are used are "integrand" and less often, "primitive."