If int frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = A | vedclass

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

Question

(A) We use the method of partial fractions to decompose the integrand: $\frac{1}{(x^{2} + 4)(x^{2} + 9)} = \frac{1}{5} \left( \frac{1}{x^{2} + 4} - \f

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) We use the method of partial fractions to decompose the integrand: $\frac{1}{(x^{2} + 4)(x^{2} + 9)} = \frac{1}{5} \left( \frac{1}{x^{2} + 4} - \frac{1}{x^{2} + 9} ight)$.Integrating both sides with respect to $x$:$\int \frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = \frac{1}{5} \int \frac{1}{x^{2} + 2^{2}} dx - \frac{1}{5} \int \frac{1}{x^{2} + 3^{2}} dx$.Using the standard formula $\int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} 	an^{-1} \frac{x}{a} + C$:$= \frac{1}{5} \left( \frac{1}{2} 	an^{-1} \frac{x}{2} ight) - \frac{1}{5} \left( \frac{1}{3} 	an^{-1} \frac{x}{3} ight) + C$.$= \frac{1}{10} 	an^{-1} \frac{x}{2} - \frac{1}{15} 	an^{-1} \frac{x}{3} + C$.Comparing this with $A 	an^{-1} \frac{x}{2} + B 	an^{-1} \frac{x}{3} + C$,we get $A = \frac{1}{10}$ and $B = -\frac{1}{15}$.Therefore,$A - B = \frac{1}{10} - (-\frac{1}{15}) = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$.

If $\int \frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = A \tan^{-1} \frac{x}{2} + B \tan^{-1} \left( \frac{x}{3} \right) + C$,then $A - B =$

Similar Questions

$\int \frac{dx}{\sin x + \sin 2x} = $

$\frac{x^4}{(x^2+1)(x^2+3)} =$

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

$\int \frac{3x-2}{(x+1)(x-2)^2} dx = $ (where $C$ is a constant of integration.)

$\int \frac{x + 2}{x^2 - 7x + 12} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module