Question bank 2013

\begin{aligned} &\text { Sol.(A) }\\ &\begin{aligned} & \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \\ & \Rightarrow y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \\ & \Rightarrow y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \\ & \underline{y(-1)}=-\frac{\pi}{4} ;-\frac{\pi}{4}=0-\frac{\pi}{4}+C \Rightarrow C=0 \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x \\ & \underline{y(0)}=\tan ^{-1} 1=\frac{\pi}{4} \end{aligned} \end{aligned}

\begin{aligned} &\text { Sol.(A) }\\ &\begin{aligned} & \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \\ & \Rightarrow y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \\ & \Rightarrow y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \\ & \underline{y(-1)}=-\frac{\pi}{4} ;-\frac{\pi}{4}=0-\frac{\pi}{4}+C \Rightarrow C=0 \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x \\ & \underline{y(0)}=\tan ^{-1} 1=\frac{\pi}{4} \end{aligned} \end{aligned}

\begin{array}{lcr} \begin{aligned} Sol.(A) \\ \begin{aligned} \int d y = \int \frac{(2 x^{2} + 2 x + 3)}{x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2} d x \\ \Rightarrow y = \int \frac{(2 x^{2} + 2 x + 3)}{(x^{2} + 1) (x^{2} + 2 x + 2)} d x \\ \Rightarrow y = \int \frac{d x}{x^{2} + 2 x + 2} + \int \frac{d x}{x^{2} + 1} \\ \Rightarrow y = \tan^{- 1} (x + 1) + \tan^{- 1} x + C \\ y (- 1) = - \frac{π}{4}; - \frac{π}{4} = 0 - \frac{π}{4} + C \Rightarrow C = 0 \\ \Rightarrow y = \tan^{- 1} (x + 1) + \tan^{- 1} x \\ y (0) = \tan^{- 1} 1 = \frac{π}{4} \end{aligned} \end{aligned} \end{array}

(1) $\begin{equation*} \begin{aligned} &\text { Sol.(A) }\\ &\begin{aligned} & \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \\ & \Rightarrow y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \\ & \Rightarrow y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \\ & \underline{y(-1)}=-\frac{\pi}{4} ;-\frac{\pi}{4}=0-\frac{\pi}{4}+C \Rightarrow C=0 \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x \\ & \underline{y(0)}=\tan ^{-1} 1=\frac{\pi}{4} \end{aligned} \end{aligned} \end{equation*}$