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

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

Question

(D) To solve the integral $I = \int \frac{dx}{(x^2+1)(x^2+4)}$,we use partial fractions. Let $x^2 = t$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) To solve the integral $I = \int \frac{dx}{(x^2+1)(x^2+4)}$,we use partial fractions. Let $x^2 = t$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$.By partial fraction decomposition,$1 = A(t+4) + B(t+1)$.Setting $t = -1$,we get $1 = 3A \implies A = \frac{1}{3}$.Setting $t = -4$,we get $1 = -3B \implies B = -\frac{1}{3}$.Thus,$\frac{1}{(x^2+1)(x^2+4)} = \frac{1}{3} \left( \frac{1}{x^2+1} - \frac{1}{x^2+4} ight)$.Integrating both sides,$I = \frac{1}{3} \int \frac{dx}{x^2+1} - \frac{1}{3} \int \frac{dx}{x^2+2^2}$.Using the standard formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \frac{1}{3} 	an^{-1} x - \frac{1}{3} \cdot \frac{1}{2} 	an^{-1}(\frac{x}{2}) + C$.$I = \frac{1}{3} 	an^{-1} x - \frac{1}{6} 	an^{-1}(\frac{x}{2}) + C$.

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

Similar Questions

Integrate the rational function: $\frac{x}{(x^{2}+1)(x-1)}$

Integrate the rational function: $\frac{x}{(x-1)^{2}(x+2)}$

Integrate the rational function: $\frac{3x+5}{x^{3}-x^{2}-x+1}$

$\int \frac{dx}{x(x^2+1)^3} = ?$

$\int \frac{dx}{x(x^3+1)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module