int frac{d x}{sqrt{5+4 x-x^{2}}}= | vedclass

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

Question

(A) To evaluate the integral $I = \int \frac{dx}{\sqrt{5+4x-x^2}}$,we first complete the square for the quadratic expression inside the square root.$5

Vedclass · Accepted Answer

$\sin ^{-1}\left(\frac{x-2}{3}\right)+c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{dx}{\sqrt{5+4x-x^2}}$,we first complete the square for the quadratic expression inside the square root.$5+4x-x^2 = -(x^2-4x-5) = -((x-2)^2 - 4 - 5) = -( (x-2)^2 - 9 ) = 9 - (x-2)^2$.Substituting this back into the integral,we get:$I = \int \frac{dx}{\sqrt{3^2 - (x-2)^2}}$.Using the standard integration formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we obtain:$I = \sin^{-1}\left(\frac{x-2}{3}\right) + c$.

$\int \frac{d x}{\sqrt{5+4 x-x^{2}}}=$

Similar Questions

The value of $\int \frac{d x}{7+6 x-x^2}$ is equal to

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

Find the following integral: $\int \operatorname{cosec} x(\operatorname{cosec} x + \cot x) \, dx$

If $\int [ \cos(x) \cdot \frac{d}{dx}(\csc(x)) ] dx = f(x) + g(x) + c$,then $f(x) \cdot g(x) =$

$\int \frac{1}{\sqrt{4x-x^2}} dx = $ . . . . . . $+ c$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module