int_{2}^{3} frac{dx}{x^{2}+x} = | vedclass

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

Question

(C) We are given the integral $I = \int_{2}^{3} \frac{dx}{x^{2}+x}$.First,we factor the denominator: $x^{2}+x = x(x+1)$.Using partial fractions,we can

Vedclass · Accepted Answer

$\log \left(\frac{9}{8}\right)$

Vedclass · Answer

(C) We are given the integral $I = \int_{2}^{3} \frac{dx}{x^{2}+x}$.First,we factor the denominator: $x^{2}+x = x(x+1)$.Using partial fractions,we can write $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$.Solving for $A$ and $B$,we get $1 = A(x+1) + Bx$. Setting $x=0$ gives $A=1$,and setting $x=-1$ gives $B=-1$.Thus,$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$.Now,integrate: $I = \int_{2}^{3} \left(\frac{1}{x} - \frac{1}{x+1}ight) dx = [\log|x| - \log|x+1|]_{2}^{3} = [\log|\frac{x}{x+1}|]_{2}^{3}$.Evaluating at the limits: $I = \log(\frac{3}{4}) - \log(\frac{2}{3}) = \log(\frac{3}{4} \div \frac{2}{3}) = \log(\frac{3}{4} 	imes \frac{3}{2}) = \log(\frac{9}{8})$.

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

Similar Questions

The value of $\int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$ is

If $\int_{\pi/6}^{\pi/4} (\cot (x - \frac{\pi}{3}) \cot (x + \frac{\pi}{3}) + 1) dx = a \log_e (\sqrt{3} - 1)$, then $9a^2$ is equal to . . . . . . .

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

If $[.]$ represents the greatest integer function,then evaluate the integral: $\int_{\frac{3 \pi}{4}}^\pi \left[ \sin x + \left[ \frac{4 x}{\pi} \right] \right] dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module