Question

Find the integral of √(x² +4) W.R.T x​

Answer 1

Answer:

$\frac{x}{2} *\sqrt{x^{2} +4}$ +$\frac{1}{2}$*LN(|$\frac{x+\sqrt{x^{2} +4} }{2}$|) +C

Step-by-step explanation:

we will have to do a trig sub for this

use x=a*tanθ for sqrt(x^2 +a^2) where a=2

x=2tanθ, dx= 2 sec^2 (θ) dθ

this turns $\int\limits {\sqrt{x^{2}+4 } } \, dx$ into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ

the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1

then it simplifies into integral(4*sec^3 (θ)) dθ

you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C

then you will need to rework your functions of θ back into functions of x

tanθ will resolve back into $\frac{x}{2}$ (see substitutions) while secθ will resolve into $\frac{\sqrt{x^{2} +4} }{2}$

sec(θ)=$\frac{\sqrt{x^{2} +4} }{2}$  is from its ratio identity of hyp/adj where the hyp. is $\sqrt{x^{2} +4}$  and adj is 2 (see tan(θ) ratio)

after resolving back into functions of x, substitute ratios for trig functions:

= $\frac{x}{2} *\sqrt{x^{2} +4}$ + $\frac{1}{2}$*LN(|$\frac{x+\sqrt{x^{2} +4} }{2}$|) +C