Question #362bd

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Answer 1 · 2016-11-18T12:51:15

$2log_e(sqrtx+1)$

Note that $1/(x+sqrt(x))=1/(sqrtx(sqrtx+1))$ and also that
$d/dx(sqrtx+1)=1/2(1/sqrtx)$ so

$1/(sqrtx(sqrtx+1))=2(d/dx(sqrtx+1))/(sqrtx+1)$ so

$intdx/(x+sqrt(x))=int1/(x+sqrt(x))dx=int2(d/dx(sqrtx+1))/(sqrtx+1)dx=2log_e(sqrtx+1)$

Answer 2 · 2016-11-18T12:53:45

The answer is $=2ln(1+sqrtx)+C$

The two writing represent the same integral.

Let $I=intdx/(x+sqrtx)$

Multiply numerator and denominator by $x-sqrtx$

$I=int((x-sqrtx)dx)/((x+sqrtx)(x-sqrtx))$

$=int((x-sqrtx)dx)/(x^2-x)$

Solving by substitution

Let $u=sqrtx$ ;
$u^2=x$

$u^4 =x^2$

$du=dx/(2sqrtx)$

$dx=2udu$

$I=int(u^2-u)(2udu)/(u^4-u^2)$

$=2int(u^2(u-1)du)/(u^2(u^2-1))$

$=2int(cancel(u-1)du)/((u+1)cancel(u-1))$

$=2ln(u+1)$

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