Evaluate the integral: int sqrt{x^2+x+1} , dx | vedclass

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

Question

(A) To evaluate $\int \sqrt{x^2+x+1} \, dx$,we complete the square inside the square root: $x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4} = (x+\frac{1}{

Vedclass · Accepted Answer

$\frac{(2x+1)}{4} \sqrt{x^2+x+1} + \frac{3}{8} \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + c$

Vedclass · Answer

(A) To evaluate $\int \sqrt{x^2+x+1} \, dx$,we complete the square inside the square root: $x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4} = (x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$. Let $u = x+\frac{1}{2}$,then $du = dx$. The integral becomes $\int \sqrt{u^2 + a^2} \, du$ where $a = \frac{\sqrt{3}}{2}$. Using the standard formula $\int \sqrt{u^2+a^2} \, du = \frac{u}{2}\sqrt{u^2+a^2} + \frac{a^2}{2}\ln|u+\sqrt{u^2+a^2}| + c$. Substituting back $u = x+\frac{1}{2}$ and $a^2 = \frac{3}{4}$: $= \frac{x+1/2}{2}\sqrt{x^2+x+1} + \frac{3/4}{2}\ln|x+\frac{1}{2} + \sqrt{x^2+x+1}| + c$. $= \frac{2x+1}{4}\sqrt{x^2+x+1} + \frac{3}{8}\sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + c$.

Evaluate the integral: $\int \sqrt{x^2+x+1} \, dx$

Similar Questions

$\int \frac{dx}{\tan x + \cot x + \sec x + \csc x} =$

Find the integral of the function $\frac{1}{\sin x \cos ^{3} x}$.

Let $\alpha \in (0, \pi /2)$ be fixed. If the integral $\int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha} dx = A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$,where $C$ is a constant of integration,then the functions $A(x)$ and $B(x)$ are respectively:

$\int \frac{d x}{\sin x+\cos x}=$

If $f(x) = \int \operatorname{cosec}^5 x \, dx$,then $f\left(\frac{\pi}{4}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module