int frac{1 + x + sqrt{x + x^2}}{sqrt{x} + sqr | vedclass

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

Question

(B) Given integral is $I = \int \frac{1 + x + \sqrt{x(1 + x)}}{\sqrt{x} + \sqrt{1 + x}} dx$. We can rewrite the numerator as: $1 + x + \sqrt{x} \cdot

Vedclass · Accepted Answer

$\frac{2}{3} (1 + x)^{3/2} + c$

Vedclass · Answer

(B) Given integral is $I = \int \frac{1 + x + \sqrt{x(1 + x)}}{\sqrt{x} + \sqrt{1 + x}} dx$. We can rewrite the numerator as: $1 + x + \sqrt{x} \cdot \sqrt{1 + x} = \sqrt{1 + x} \cdot \sqrt{1 + x} + \sqrt{x} \cdot \sqrt{1 + x}$. Factoring out $\sqrt{1 + x}$,we get: $\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})$. Substituting this back into the integral: $I = \int \frac{\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})}{\sqrt{x} + \sqrt{1 + x}} dx$. Canceling the common term $(\sqrt{x} + \sqrt{1 + x})$: $I = \int \sqrt{1 + x} dx = \int (1 + x)^{1/2} dx$. Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + c$: $I = \frac{(1 + x)^{3/2}}{3/2} + c = \frac{2}{3} (1 + x)^{3/2} + c$.

$\int \frac{1 + x + \sqrt{x + x^2}}{\sqrt{x} + \sqrt{1 + x}} dx =$

Similar Questions

$\int \sqrt{1 + \sin \frac{x}{2}} \, dx = $

If $x > 0$ and $x \neq (2n+1) \frac{\pi}{2}$, then $\int \left(x \sqrt{x} - e^{\log(\sec x \tan x)} + \frac{3x^2 - 2x + 1}{x^2}\right) dx =$

$\int {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx} = $

Find the following integral: $\int(4 e^{3 x}+1) d x$

$ \int \frac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} dx $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module