How do you find the integral of $\int \frac{1}{\sqrt{x} \cdot (1 + x)} d x$ $int 1/(sqrt(x)*(1+x)) dx$ ?

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How do you find the integral of $\int \frac{1}{\sqrt{x} \cdot (1 + x)} d x$ $int 1/(sqrt(x)*(1+x)) dx$ ?

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Answer 1 · 2018-07-09T03:49:41

$\int \frac{1}{\sqrt{x} (1 + x)} d x = 2 a r c tan (\sqrt{x}) + c$ $int1/(sqrtx(1+x))dx=2arc tan(sqrtx)+c$

Explanation:

We know that,

$(1) \frac{d}{d x} (a r c tan x) = \frac{1}{1 + x^{2}} \Rightarrow \int \frac{1}{1 + x^{2}} d x = a r c tan x + c$ $color(red)((1)d/(dx)(arc tanx)=1/(1+x^2)=>int1/(1+x^2)dx=arc tanx+c$

We have ,

$I = \int \frac{1}{\sqrt{x} (1 + x)} d x$ $I=int1/(sqrtx(1+x))dx$

Subst. $\sqrt{x} = u \Rightarrow x = u^{2} \Rightarrow d x = 2 u d u$ $color(blue)(sqrtx=u=>x=u^2=>dx=2udu$

So ,

$I = \int \frac{1}{u (1 + u^{2})} 2 u \cdot d u$ $I=int1/(u(1+u^2))2u*du$

$=2int1/(1+u^2)du...tocolor(red)(Apply(1)$

$=2arc tan u+c$

Subst. back $color(blue)(u=sqrtx$ ,we get

$I=2arc tan(sqrtx)+c$

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How do you find the integral of $\int \frac{1}{\sqrt{x} \cdot (1 + x)} d x$ $int 1/(sqrt(x)*(1+x)) dx$ ?

1 Answer

Explanation:

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How do you find the integral of int 1/(sqrt(x)*(1+x)) dx∫1√x⋅(1+x)dxint 1/(sqrt(x)*(1+x)) dx?

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How do you find the integral of $\int \frac{1}{\sqrt{x} \cdot (1 + x)} d x$ $int 1/(sqrt(x)*(1+x)) dx$ ?