integration question

[M-D]

I have the following question and i have done the following steps:













that is what i get. however the solutions have got


i don't understand where comes from. your help is greatly appreciated.

[silverpixeli]

short answer, the number you left in the numerator of your integrand before you integrated to an arctan function was wrong, arctan(x/b) integrates to b/(x^2 + b^2) and you have it in the form of (b^2)/(x^2 + b^2) (the difference between the two being the 'b squared' v just 'b')
you need to take out an extra b before you do the integration

Heres the correct working incase you get stuck

Spoiler

[Jeggz]

In step number 3, you need to try and get 'a' on the numerator, not If that makes sense...

Because to recognise it is an arctan(, we need to get it in the form . If you replace the numerator with a, which in this case will equal then you will get the right answer. Hope that makes sense 

EDIT: Beaten 

[M-D]

thanks silverpixeli and Jeggz. 

[M-D]

could i get some help with the following 3 questions also:

1) differentiate

2)

3)

all these are meant to be done without a calculator. i don't know hoe to do them. even if someone could give a hint on them it will be much appreciated. thanks

[brightsky]

1. just use the chain rule. d/dx(arccos(f(x))) = -1/sqrt(1-(f(x))^2) f'(x)
2. use linear substitution. let u = x-4, du=dx. you know then that x = u+4, so 3x+2 = 3(u+4) + 2 = 3u + 14. so the integral becomes int (3u+14)u^1/2 du. expand the integrand and then integrate term by term.
3. x^2/(9+x^2) = (9+x^2 - 9)/(9+x^2) = 1 - 9/(9+x^2). find the integral of that. the second part would be arctan(something).

[M-D]

thanks brightsky. how do you actually figure out how to start solving questions like these? is there anything particular you look for? thanks.

[lzxnl]

The first one is just chain rule.
For the second one, it's not nice to have linear multiples of linear square roots. It would be much nicer to have linear term multiplied by sqrt(u), which you could then expand easily. That's why we make the linear substitution.
For the third one, simplify whenever the degree of the top is equal to or higher than the degree of the bottom. Then you should have an idea.