The value of int_{0}^{1} frac{2x^2 + 3x + 3}{ | vedclass

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

Question

(D) Let $I = \int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$. Using partial fractions,we write $\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) Let $I = \int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$. Using partial fractions,we write $\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}$. Equating coefficients,we get $A(x^2+2x+2) + (Bx+C)(x+1) = 2x^2+3x+3$. For $x = -1$,$A(1-2+2) = 2-3+3 \implies A = 2$. Comparing coefficients of $x^2$,$A+B = 2 \implies B = 0$. Comparing constants,$2A+C = 3 \implies 4+C = 3 \implies C = -1$. So,$I = \int_{0}^{1} (\frac{2}{x+1} - \frac{1}{x^2+2x+2}) dx = \int_{0}^{1} (\frac{2}{x+1} - \frac{1}{(x+1)^2+1}) dx$. $I = [2 \ln|x+1| - 	an^{-1}(x+1)]_{0}^{1} = (2 \ln 2 - 	an^{-1} 2) - (2 \ln 1 - 	an^{-1} 1) = 2 \ln 2 - 	an^{-1} 2 + \frac{\pi}{4}$. Since $	an^{-1} 2 + \cot^{-1} 2 = \frac{\pi}{2}$,we have $\frac{\pi}{4} - 	an^{-1} 2 = \frac{\pi}{4} - (\frac{\pi}{2} - \cot^{-1} 2) = \cot^{-1} 2 - \frac{\pi}{4}$. Thus,$I = 2 \ln 2 + \cot^{-1} 2 - \frac{\pi}{4}$. Both options $A$ and $B$ are equivalent. Therefore,the correct answer is $D$.

The value of $\int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$ is:

Similar Questions

Evaluate the definite integral $\int_{-1}^{2} \left[ \frac{[x]}{1 + x^2} \right] dx$,where $[\cdot]$ denotes the Greatest Integer Function $(GIF)$.

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

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$, then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

Let $f:[0,1] \rightarrow [0, \infty)$ be a continuous function such that $\int_0^1 f(x) dx = 10$. Which of the following statements is $NOT$ necessarily true?

If $\int\limits_0^a {\frac{{dx}}{{\sqrt {x + a} + \sqrt x }}} = \int\limits_0^{\frac{\pi }{8}} {\frac{{2\tan \theta }}{{\sin 2\theta }}} d\theta$,then the value of $a$ is equal to $(a > 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module