What is the antiderivative of $(2x)/(sqrt(5 + 4x))$ ?

Question

1599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-23T12:36:18

$I=1/8{5+4x-10lnsqrt(5+4x)}+C$ , or,

$I=1/4(2x-5lnsqrt(5+4x))+K$ , where, $K=5/8+C$ .

We take substn. $5+4x=t^2, so that, 4dx=2tdt, i.e., dx=1/2dt$ .

Also, $5+4x=t^2 rArr x=(t^2-5)/4$ . Therefore,

$I=int(2x)/sqrt(5+4x)dx=1/2int{2(t^2-5)/4}/sqrt(t^2)dt=1/4int(t^2-5)/tdt$

$=1/4int(t^2/t-5/t)dt=1/4(t^2/2-5ln|t|)$

$=1/8(t^2-10ln|t|)$ . Therefore,

$I=1/8{5+4x-10lnsqrt(5+4x)}+C$

Readjusting the Constant of Integration, we may like,

$I=1/4(2x-5lnsqrt(5+4x))+K$ , where, $K=5/8+C$ .

Enjoy Maths.!

What is the antiderivative of (2x)/(sqrt(5 + 4x))(2x)/(sqrt(5 + 4x))?