Question

integrate this question ​

Answer 1

AnswEr :

Given Expression,

$\displaystyle \sf l = \int \dfrac{dx}{ \sqrt{x} + x}$

The expression can be re-written as :

$\implies \: \displaystyle \sf l = \int \dfrac{dx}{ \sqrt{x}(1 + \sqrt{x} )}$

Let us assume u = √x + 1.

Differentiating w.r.t x on both sides,

$\sf \dfrac{du}{dx} = \dfrac{1}{2 \sqrt{x} } \\ \\ \longrightarrow \: \sf \: dx = 2 \sqrt{x}.du$

Now,

$\implies \: \displaystyle \sf l = \int \dfrac{1}{ \cancel{ \sqrt{x}}(u )} \times 2 \cancel{\sqrt{x}} du \\ \\ \implies \: \displaystyle \sf l =2 \int \dfrac{1}{u }du \\ \\ \implies \: \sf \: l = 2 \times log(u) \\ \\ \implies \: \sf \: l = 2log(1 + \sqrt{x} ) + c$

Answer 2

Question:

$I = ∫ \frac{dx}{ \sqrt{x} + x }$

Answer:

$\boxed{I = 2log(1 + \sqrt{x} ) + c}$

Step-by-step explanation:

$I = ∫ \frac{dx}{ \sqrt{x} + x }$

$\longrightarrow I = ∫ \frac{dx}{ \sqrt{x}(1 + \sqrt{x}) }$

$\underline\bold{u = \sqrt{x} + 1}$

$\longrightarrow \frac{du}{dx} = \frac{1}{2 \sqrt{x} }$

$\longrightarrow \underline{dx = 2 \sqrt{x} \times du}$

Thus,

$I = ∫ \frac{1}{ \sqrt{x}(u) } \times 2 \sqrt{x} du$

$= 2 \times ∫ \frac{1}{u} du$

$= 2 \times log(u)$

$\longrightarrow\boxed{I = 2log(1 + \sqrt{x} ) + c}$