int frac{2 x^2-1}{left(x^2+4right)left(x^2-3r | vedclass

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

Question

(A) Let $I = \int \frac{2x^2-1}{(x^2+4)(x^2-3)} dx$. Using partial fractions,let $t = x^2$. Then $\frac{2t-1}{(t+4)(t-3)} = \frac{A}{t+4} + \frac{B}{t

Vedclass · Accepted Answer

$\frac{9}{14} 	an ^{-1}\left(\frac{x}{2}ight)+\frac{5}{14 \sqrt{3}} \log \left|\frac{x-\sqrt{3}}{x+\sqrt{3}}ight|+c$

Vedclass · Answer

(A) Let $I = \int \frac{2x^2-1}{(x^2+4)(x^2-3)} dx$. Using partial fractions,let $t = x^2$. Then $\frac{2t-1}{(t+4)(t-3)} = \frac{A}{t+4} + \frac{B}{t-3}$. $2t-1 = A(t-3) + B(t+4)$. For $t = 3$,$5 = 7B \Rightarrow B = \frac{5}{7}$. For $t = -4$,$-9 = -7A \Rightarrow A = \frac{9}{7}$. Thus,$I = \int \left( \frac{9/7}{x^2+4} + \frac{5/7}{x^2-3} ight) dx$. $I = \frac{9}{7} \int \frac{1}{x^2+2^2} dx + \frac{5}{7} \int \frac{1}{x^2-(\sqrt{3})^2} dx$. Using standard integrals $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a})$ and $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log |\frac{x-a}{x+a}|$: $I = \frac{9}{7} \cdot \frac{1}{2} 	an^{-1}(\frac{x}{2}) + \frac{5}{7} \cdot \frac{1}{2\sqrt{3}} \log |\frac{x-\sqrt{3}}{x+\sqrt{3}}| + c$. $I = \frac{9}{14} 	an^{-1}(\frac{x}{2}) + \frac{5}{14\sqrt{3}} \log |\frac{x-\sqrt{3}}{x+\sqrt{3}}| + c$.

$\int \frac{2 x^2-1}{\left(x^2+4\right)\left(x^2-3\right)} d x=$

Similar Questions

$\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x$ is

If $\int \frac{x}{\left(x^2+1\right)(x-1)} d x=A \log \left|x^2+1\right|+B \tan ^{-1} x+C \log |x-1|+d$, then $A+B+C=$

$\int \frac{d x}{(\sin x+\cos x)(2 \cos x+\sin x)} = $

$\int \frac{dx}{x^3+3x^2+2x} = $

The value of $\int_1^2 \frac{dx}{x(1 + x^4)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module