int sqrt{x^{2}+2 x+5} , dx is equal to | vedclass

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

Question

(A) We need to evaluate the integral $ I = \int \sqrt{x^{2}+2 x+5} \, dx $. First,complete the square for the quadratic expression: $ x^{2}+2x+5 = (x+

Vedclass · Accepted Answer

$ \frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+C $

Vedclass · Answer

(A) We need to evaluate the integral $ I = \int \sqrt{x^{2}+2 x+5} \, dx $. First,complete the square for the quadratic expression: $ x^{2}+2x+5 = (x+1)^{2} + 4 = (x+1)^{2} + 2^{2} $. Now the integral becomes $ I = \int \sqrt{(x+1)^{2} + 2^{2}} \, dx $. Using the standard formula $ \int \sqrt{t^{2}+a^{2}} \, dt = \frac{t}{2} \sqrt{t^{2}+a^{2}} + \frac{a^{2}}{2} \log \left|t + \sqrt{t^{2}+a^{2}}\right| + C $,where $ t = x+1 $ and $ a = 2 $. Substituting these values: $ I = \frac{x+1}{2} \sqrt{(x+1)^{2} + 2^{2}} + \frac{2^{2}}{2} \log \left|(x+1) + \sqrt{(x+1)^{2} + 2^{2}}\right| + C $. Simplifying the expression: $ I = \frac{1}{2}(x+1) \sqrt{x^{2}+2x+5} + 2 \log \left|x+1 + \sqrt{x^{2}+2x+5}\right| + C $. Thus,the correct option is $ A $.

$ \int \sqrt{x^{2}+2 x+5} \, dx $ is equal to

Similar Questions

For $-1 < x, y < 1$, if $\int \frac{x}{\sqrt{1+x}+\sqrt{1-x}} dx + \int \frac{y}{\sqrt{y+1}+\sqrt{y-1}} dy = A(1+x)^{3/2} + B(1-x)^{3/2} + f(y)(y+1)^{3/2} + g(y)(y-1)^{3/2} + C$, then $A f(y) + B g(y) =$

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

$\int \sec x \tan^3 x \, dx = $

$\int \left[ \log (1+\cos x) - x \tan \left( \frac{x}{2} \right) \right] dx =$

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + k$ where $k$ is a constant of integration,then $A+B+C$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module