Question

Integral question
I have to integrate this function (20x - 5x^2) / (x^2 + 9) between 4 and 0.

The final answer has to be 20ln(5/3) + 15tan-1(4/3) -20.

However, my calculator, even after somehow simplifying it a bit gives this:
( 40ln(5/3) -30tan-1(3/4) +15pi -40) / 2.

As you can see there is something wrong with arctan integration, can anybody help or know how to simplify this with some identity?

Answer 1

I'm asking for this because this is one of my practise exams, and in the exam i will have to use this calculator, and if i get a similar answer again, i will not have a clue of how to simplify it?

Answer 2

Simplify \(\dfrac{20 - 5x^2}{x^2 + 9}\) first

Answer 3

I know, partial fractions et all, but in the exam I don't think I would have time for that... ;/

Answer 4

\(-\dfrac{5x^2 - 20}{x^2 + 9} = -\dfrac{5x^2 + 45 - 65}{x^2 + 9} = -\dfrac{5x^2 + 45}{x^2 + 9} + \dfrac{65}{x^2 + 9} = -5 + \dfrac{65}{x^2 + 9}\)

Answer 5

You don't have to know partial fractions to do this

Answer 6

You know that \(5(x^2 + 9) = 5x^2 + 45\)

Answer 7

Use that to help you simplify it

Answer 8

Now you can easily integrate this:

\[\int -5 dx + \int \frac{65}{x^2 + 9} dx\]

Answer 9

actually, i came across this at the other forums
tan−1(3/4) +tan−1(4/3)=π/2 and it seems to work for any numbers. Do you know what this identity is?

Answer 10

If you want integration rules, use these

Answer 11

http://myhandbook.info/form_integ.html

Answer 12

From that list, you can apply #8

Answer 13

I solved it but now the ln part disappears pretty much all together. And I still get the answer of tan-1(3/4) value instead of tan-1(4/3)

Answer 14

I know how to integrate to get inverse tan functions. But from what you did, the ln part for my integral disappeared all together. Also, as I was saying, I do get the right answer, and also just different tan-1, which is weird. But then this sort of identity thing solves my initial question... I just don't know what it is. But thanks :)