The primitive of (x^2 + 4)^{-1/2} with respec | vedclass

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

Question

(C) Let $u = x^2 + 2$. Then $du = 2x \, dx$. We want to find the integral $\int (x^2 + 4)^{-1/2} \, d(x^2 + 2)$. Since $d(x^2 + 2) = 2x \, dx$,the int

Vedclass · Accepted Answer

$2\sqrt{x^2 + 4} + C$

Vedclass · Answer

(C) Let $u = x^2 + 2$. Then $du = 2x \, dx$. We want to find the integral $\int (x^2 + 4)^{-1/2} \, d(x^2 + 2)$. Since $d(x^2 + 2) = 2x \, dx$,the integral becomes $\int \frac{1}{\sqrt{x^2 + 4}} \cdot 2x \, dx$. Let $t = x^2 + 4$. Then $dt = 2x \, dx$. The integral becomes $\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt$. Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$,we get: $\frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$. Substituting back $t = x^2 + 4$,we get $2\sqrt{x^2 + 4} + C$.

The primitive of $(x^2 + 4)^{-1/2}$ with respect to $x^2 + 2$ is equal to :-

Similar Questions

If $\int_1^3 x^n \sqrt{x^2-1} dx = 6$,then $n = $

$\int \frac{a \, dx}{b + c e^x} = $

If $\int(7x-2)\sqrt{3x+2} \, dx = A(3x+2)^{\frac{5}{2}} + B(3x+2)^{\frac{3}{2}} + c$ (where $c$ is a constant of integration),then the values of $A$ and $B$ are respectively:

If $\int {\frac{{\left( {2x + 3} \right)dx}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 1}}} = C - \frac{1}{{f(x)}}$ where $f(x)$ is of the form of $ax^2 + bx + c$ then $(a + b + c)$ equals

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module