# Issues With Simple Integration Problem

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1. Jan 15, 2014

### 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.

1. The problem statement, all variables and given/known data
Evaluate: $$\int \frac{(t+9)^2}{t^3} dx$$

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

3. 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
$$= ln(t) - \frac{18}{t} - \frac{81}{2t^2} + c$$
$$\int \frac{(t+9)^2}{t^3} dx = ln(t) - \frac{18}{t} - \frac{81}{2t^2} + c$$

2. Jan 15, 2014

### Differentiate1

You can use WolframAlpha to check it if you're unsure of any answer you have in the future.

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

3. Jan 15, 2014

### 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.

4. Jan 15, 2014

### 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

5. Jan 16, 2014

### Staff: Mentor

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."