Question

\[\int\limits_{1}^{4}{1 \over \sqrt{x}+x}dx\]

Answer 1

multiply top and bottom by sqrt(x)-x i think..

Answer 2

let x be tan^2 \(\theta\)
dx= 2 tan \(\theta\) sec^2 \(\theta\)
when x=1
tan^2 \(\theta\)= 1
\(\theta\)= \(\pi/4\)
when x=4
tan^2 \(\theta\)=4
\(\theta\)= \(\tan^{-1} 2\)
so we have the integral now
\[\int_{\theta=\pi/4}^{\theta=\ tan^{-1} 2} \frac{ 2 tan (\theta ) sec^2 (\theta)} { tan \theta+ tan^2 \theta } d\theta \]
now let's cancel the tan \(\theta\) from numerator and denominator
we get
\[\int_{\theta=\pi/4}^{\theta=\ tan^{-1} 2} \frac{ 2 sec^2 (\theta)} { tan \theta+ 1}
d\theta \]
let's substitute ( 1+ tan \(\theta\)) as t
so we get
sec^2 \(\theta\) d\(\theta\)= dt
now when \(\theta\)= \(\pi/4\)
t= 2
and when \(\theta\)= \(\tan^{-1} 2\)
t= 3
so now we have integral as
\[ \int_{t=2} ^{t=3} \frac{2}{t} dt\]
we know that integral of 1/t is log t (with base e)
so we get
\[ [2 \log_{e} t]_{t=2}^{t=3} \]
we get finally
\[ 2 \log_{e} 3-2 \log_{e} 2\]
or
\[2\log_{e} {\frac{3}{2}}\]