int frac{dx}{sqrt{x}+x} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To solve the integral $I = \int \frac{dx}{\sqrt{x} + x}$,we can simplify the denominator by factoring out $\sqrt{x}$.$I = \int \frac{dx}{\sqrt{x}(

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) To solve the integral $I = \int \frac{dx}{\sqrt{x} + x}$,we can simplify the denominator by factoring out $\sqrt{x}$.$I = \int \frac{dx}{\sqrt{x}(1 + \sqrt{x})}$Let $u = 1 + \sqrt{x}$. Then,$du = \frac{1}{2\sqrt{x}} dx$,which implies $\frac{dx}{\sqrt{x}} = 2 du$.Substituting these into the integral,we get:$I = \int \frac{2 du}{u} = 2 \int \frac{du}{u} = 2 \log |u| + c$.Substituting back $u = 1 + \sqrt{x}$,we obtain:$I = 2 \log (1 + \sqrt{x}) + c$.

$\int \frac{dx}{\sqrt{x}+x} = $

Similar Questions

$\int \frac{x^3}{\sqrt{1 - x^8}} \, dx = $

Integrate the function $\frac{\sin x}{1+\cos x}$.

Let $\int \frac{x^{1/2}}{\sqrt{1-x^3}} dx = \frac{2}{3} g(f(x)) + c$; then

$ \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $ is equal to

If $f(x) = \int \frac{16x^7 + 5x^{10}}{(x^3 + 2 + 3x^8)^2} dx$ for $x \geq 0$ and $f(0) = 1$, then the value of $f(1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module