Evaluate the following integral: int_{1}^{2} | vedclass

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

Question

(B) Let $I = \int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$.Using partial fractions,we decompose the integrand:$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac

Vedclass · Accepted Answer

$\log \left( \frac{32}{27} \right)$

Vedclass · Answer

(B) Let $I = \int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$.Using partial fractions,we decompose the integrand:$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$x = A(x+2) + B(x+1)$Setting $x = -1$,we get $A = -1$. Setting $x = -2$,we get $B = 2$.So,$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}$.The indefinite integral is:$\int \left( \frac{-1}{x+1} + \frac{2}{x+2} ight) dx = -\log |x+1| + 2 \log |x+2| = F(x)$.Applying the Fundamental Theorem of Calculus:$I = F(2) - F(1)$$I = [-\log(3) + 2 \log(4)] - [-\log(2) + 2 \log(3)]$$I = -\log(3) + 2 \log(4) + \log(2) - 2 \log(3)$$I = 2 \log(4) + \log(2) - 3 \log(3)$$I = \log(16) + \log(2) - \log(27)$$I = \log \left( \frac{16 	imes 2}{27} ight) = \log \left( \frac{32}{27} ight)$.

Evaluate the following integral: $\int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$

Similar Questions

If $\int_{0}^{a} \frac{dx}{1+4x^{2}} = \frac{\pi}{8}$,then $a =$

The integral $\int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$ is equal to

Let $f(x) = x - [x],$ for every real number $x,$ where $[x]$ is the integral part of $x.$ Then $\int_{-1}^{1} f(x) \, dx =$

The value of the integral $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is

If $\int_1^n [x] dx = 120$,then $n = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module