How do you integrate $\int \frac{1}{\sqrt{x} {(\sqrt{x} + 1)}^{2}}$ $int 1/(sqrtx(sqrtx+1)^2)$ using substitution?

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Answer 1 · 2016-12-30T12:33:59

The answer is $= - \frac{2}{\sqrt{x} + 1} + C$ $=-2/(sqrtx+1)+C$

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C (n!=-1)$

We use the substitution

$u = \sqrt{x} + 1$ $u=sqrtx+1$

$d u = \frac{d x}{2 \sqrt{x}}$ $du=dx/(2sqrtx)$

Therefore,

$\int \frac{d x}{\sqrt{x} {(1 + \sqrt{x})}^{2}} = 2 \int \frac{d u}{u^{2}}$ $intdx/(sqrtx(1+sqrtx)^2)=2int(du)/u^2$

$= 2 \cdot - \frac{1}{u}$ $=2*-1/u$

$= - \frac{2}{\sqrt{x} + 1} + C$ $=-2/(sqrtx+1)+C$

How do you integrate int 1/(sqrtx(sqrtx+1)^2)∫1√x(√x+1)2int 1/(sqrtx(sqrtx+1)^2) using substitution?