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

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

Question

(B) Given integral: $I = \int {\frac{{1 + x + \sqrt {x(1 + x)} }}{{\sqrt x + \sqrt {1 + x} }}dx}$ Factor the numerator: $1 + x + \sqrt x \sqrt {1 + x}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given integral: $I = \int {\frac{{1 + x + \sqrt {x(1 + x)} }}{{\sqrt x + \sqrt {1 + x} }}dx}$ Factor the numerator: $1 + x + \sqrt x \sqrt {1 + x} = \sqrt {1 + x} (\sqrt {1 + x} + \sqrt x)$ Substitute this into the integral: $I = \int {\frac{{\sqrt {1 + x} (\sqrt {1 + x} + \sqrt x )}}{{\sqrt x + \sqrt {1 + x} }}dx}$ Cancel the common term $(\sqrt {1 + x} + \sqrt x)$: $I = \int {\sqrt {1 + x} \,dx}$ Integrate using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$: $I = \frac{(1 + x)^{3/2}}{3/2} + c = \frac{2}{3}(1 + x)^{3/2} + c$.

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

Similar Questions

$\int a^x \, da = $

$\int (\tan x - \cot x)^2 \, dx = $

If $\int \frac{x}{x \tan x+1} \, dx = \log f(x) + k$,then $f\left(\frac{\pi}{4}\right) =$

If $f'(x) = \frac{1}{x} + x$ and $f(1) = \frac{5}{2}$,then $f(x) = $

$\int (1 + 2x + 3x^2 + 4x^3 + \dots) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module