int frac{dx}{4x^2 + 9} = | vedclass

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

Question

(C) We need to evaluate the integral $I = \int \frac{dx}{4x^2 + 9}$.Factor out $4$ from the denominator:$I = \int \frac{dx}{4(x^2 + \frac{9}{4})} = \f

Vedclass · Accepted Answer

$\frac{1}{6} 	an^{-1} \left( \frac{2x}{3} ight) + c$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int \frac{dx}{4x^2 + 9}$.Factor out $4$ from the denominator:$I = \int \frac{dx}{4(x^2 + \frac{9}{4})} = \frac{1}{4} \int \frac{dx}{x^2 + (\frac{3}{2})^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,where $a = \frac{3}{2}$:$I = \frac{1}{4} \cdot \frac{1}{3/2} 	an^{-1}(\frac{x}{3/2}) + c$.$I = \frac{1}{4} \cdot \frac{2}{3} 	an^{-1}(\frac{2x}{3}) + c$.$I = \frac{1}{6} 	an^{-1}(\frac{2x}{3}) + c$.

$\int \frac{dx}{4x^2 + 9} = $

Similar Questions

If $f\left( \frac{x - 4}{x + 2} \right) = 2x + 1$ for $x \in R \setminus \{ -2 \}$,then $\int f(x) \,dx$ is equal to (where $C$ is a constant of integration)

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

$\int \frac{e^x-1}{e^x+1} dx =$

$\int \frac{x \, dx}{x^2 + 4x + 5} = $

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: $\int \sqrt{x-3} \left(\sin^{-1}(\log x) + \cos^{-1}(\log x)\right) dx = \frac{\pi}{3}(x-3)^{3/2} + c$
Reason $(R)$: $\sin^{-1}(f(x)) + \cos^{-1}(f(x)) = \frac{\pi}{2}$ for $|f(x)| \le 1$
Choose the correct option:

Vedclass Test Series

Exam Paper Generator

Online Exam Module