How do you find the indefinite integral of $\int \frac{1}{1 + \sqrt{2 x}}$ $int 1/(1+sqrt(2x))$ ?

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Answer 1 · 2016-11-19T06:18:57

The answer is $(1 + \sqrt{2 x}) - ln (1 + \sqrt{2 x}) + C$ $(1+sqrt(2x))-ln(1+sqrt(2x))+C$

We do this integral by substitution

Let $u = 1 + \sqrt{2} \sqrt{x}$ $u=1+sqrt(2)sqrtx$

$\sqrt{x} = \frac{u - 1}{\sqrt{2}}$ $sqrtx=(u-1)/sqrt2$

Then, $d u = \sqrt{2} \cdot \frac{d x}{2 \sqrt{x}} = \frac{d x}{\sqrt{2} \sqrt{x}}$ $du=sqrt2*dx/(2sqrtx)=dx/(sqrt2sqrtx)$

$d x = \sqrt{2} d u \cdot \sqrt{x} = \sqrt{2} \cdot d u \cdot \frac{u - 1}{\sqrt{2}}$ $dx=sqrt2du*sqrtx=sqrt2*du*(u-1)/sqrt2$

$d x = (u - 1) d u$ $dx=(u-1)du$

Therefore,

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

$= \int (1 - \frac{1}{u}) d u = u - ln u$ $=int(1-1/u)du=u-lnu$

$\int \frac{d x}{1 + \sqrt{} 2 x} = (1 + \sqrt{2 x}) - ln (1 + \sqrt{2 x}) + C$ $intdx/(1+sqrt()2x)=(1+sqrt(2x))-ln(1+sqrt(2x))+C$

How do you find the indefinite integral of int 1/(1+sqrt(2x))∫11+√2xint 1/(1+sqrt(2x))?