Question

integral below

Answer 1

help

Answer 2

HINT: \(\displaystyle\int \dfrac{1}{4+3x^2} dx = \dfrac{1}{4}\int\dfrac{1}{1+\dfrac{3}{4}x^2}dx\)

Let \(u^2 = \dfrac{3}{4}x^2 \Rightarrow u = \dfrac{\sqrt{3}}{2}x\), \(du = \dfrac{\sqrt{3}}{2}dx\).

Hope this help.

Answer 3

Remember;
\[\int\dfrac{1}{1+x^2}dx = \tan^{-1}x + C\]

Answer 4

Can you handle it now? @earslan100

Answer 5

is the answer 1/2 arctan (x/2) +C

Answer 6

@geerky42

Answer 7

or 1/2 arctan (sqrt3/4)/2 +c

Answer 8

Not quite. Do substitiution with u first. You should get \(\displaystyle\dfrac{1}{2\sqrt{3}}\int\dfrac{1}{1+u^2}du\)