Question

Can someone help me with this integral question?

Answer 1

\[\int\limits_{0}^{2\sqrt{3}}\frac{ 3 }{ 4+x ^{2} }\]

Answer 2

Rewrite it and try bringing it to the form of a tan^(-1)(alpha)

Answer 3

\[\Large 3 \int_a^b\frac{1}{4+x^2} dx\]
Now I recommend you let 2u=x

Answer 4

\[\Large\frac{3}{4} \int \frac{1}{1+u^2}dx\]
The dx is still important here, you need to substitute it.
\[\Large \frac{dx}{du}=2 \]

Answer 5

why did you let dx/du=2?

Answer 6

Whenever you perform a substitution, you also need to substitute your dimension of integration, in the original integral we were integrating with respect to dx, but I made a substitution, therefore I also need to substitute dx.

Basically, if you substitute something for x, you want to substitute something for dx too, vaguely speaking.

Answer 7

I said:

\[\Large2u=x \]
With the reason so I can factor out a 4 in the denominator, then after performing that step, I would have the normal form of an arctangent function.

I differentiate the substitution above:
\[\Large x=2u \\
\Large \frac{dx}{du}=2 \\ \Large \therefore \ dx=2du\]

Answer 8

so do i sub 2 back in for u? i'm not sure of what i'm supposed to do

Answer 9

\[\Large \int \frac{1}{4+(\underbrace{2u}_x)^2} \underbrace{2du}_{dx} \]

Answer 10

\[\Large3 \int \frac{1}{4+(\underbrace{2u}_x)^2} \underbrace{2du}_{dx} \]

Correction* I forgot the 3 above.