Problems finding the correct antiderivative

Question

I am currently working on a problem of the course MIT 18.01SC Single Variable Calculus. The original problem can be found here (you don't need to watch it, it's just the source).

The part I'm having trouble understanding is finding the following antiderivative:

$\int_0^{10} \frac{1500}{100 + (t-5)^2} - 7 dt$

The answer is given in the video, it's $\frac{1}{10} \Big( 150 \, \text{arctan}(\frac{t-5}{10}) - 7t \Big) \vert_{1}^{10}$.

However, I can't fully determine the steps needed to get to this solution. This is how far I get on my own:

1. Substitution, $u = t - 5, du = 1 dx$

2. This gives: $\int_{-5}^{5} \frac{1500}{100 + u^2} - 7 du = \int_{-5}^{5} 1500 \frac{1}{100 + u^2} - 7 du$

3. The antiderivative of $\frac{1}{1 + x^2} = \arctan x + c$

Questions:

• How do I get the integrand into the form $\frac{1}{1 + u^2}$? When dividing by $100$ I would get $\frac{1}{1 + \frac{u^2}{100}}$ which does not help either

• What further steps are missing?

Answer 1

Factoring out 100 in the denominator actually does help because you can rewrite 100 as 10^2 and then perform another substitution z=u/10. Now you can use your known antiderivative. Edit:

Answer 2

Substitute u/10 = tanx to get the desired form.

Answer 3

When I see this problem, what I see is a number in the numerator divided by a number + a variable squared. That reminds me of the derivative of arctan. To bring it to that form, what you need is a 1 in the denominator. How do you do that? Just divide the numerator and denominator by 100. Next, try to think what you can substitute to get the square thingy in the denominator and it should be a piece of cake. ;)