How do I integrate $intdx/(sqrt(x)(sqrt(x)+1))$ ?

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Answer 1 · 2017-10-29T03:27:12

$int dx/[sqrtx*(sqrtx+1)]=2Ln(sqrtx+1)+C$

$I=int dx/[sqrtx*(sqrtx+1)]$

After using $x=(tanu)^4$ and $dx=4(tanu)^3*(secu)^2*du$ transforms, $I$ became,

$I=int (4(tanu)^3*(secu)^2*du)/[(tanu)^2*(secu)^2]$

= $int 4tanu*du$

= $4ln(secu)+C$

= $2Ln[(secu)^2]+C$

= $2Ln[(tanu)^2+1]+C$

After using $x=(tanu)^4$ and $(tanu)^2=sqrt(x)$ inverse transforms, I found,

$I=int dx/[sqrtx*(sqrtx+1)]$

= $2Ln(sqrtx+1)+C$

Note: This integral also can be found by $sqrt(x)=u$ , $x=u^2$ and $dx=2u*du$ transforms,

Answer 2 · 2017-10-29T03:35:41

$2ln(sqrtx+1)+C$

Use the substitutions:

$u = sqrtx + 1$

$du = dx/(2sqrtx)$

$intdx/(sqrtx(sqrtx+1)) = 2intdx/(2sqrtx(sqrtx+1)) = 2int(du)/u = 2ln|u|+C$

$= 2ln(sqrtx+1)+C$

Final Answer

How do I integrate intdx/(sqrt(x)(sqrt(x)+1))intdx/(sqrt(x)(sqrt(x)+1))?