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# Evaluate the following $\large\int\frac{(x^2+2)}{x+1}dx$

$\begin{array}{1 1} (A)\;I=\frac{x^2}{2}+x+3\; log |x+1|+c \\ (B)I=\frac{x^2}{2}-x+3\; log |x+1|+c \\(C)\;I=\frac{x^2}{2}-x+3\; log |x+1|-c \\ (D)\;I=-\frac{x^2}{2}-x+3\; log |x+1|+c\end{array}$

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Toolbox:
• A rational expression is said to be improper if the degree of the numerator is greater than or equal to the degree of the denominator
• $\large\int \frac{dx}{(x+a)}$$=log |x+a|+c Let \large\int\frac{(x^2+2)}{x+1}dx \large\frac{x^2+2}{x+1} is an improper rational expression, To make it a proper rational expression, divide \large\frac{x^2+2}{x+1} Therefore \large\frac{(x^2+2)}{x+1}=(x-1)+\frac{3}{x+1} substituting this in I we get Hence I=\int (x-1)dx+\int \large\frac{3}{x+1}dx =>\int xdx-\int 1dx+3 \int \frac{dx}{x+1} on integrating we get, I=\large\frac{x^2}{2}$$-x+3\; log |x+1|+c$
answered Apr 11, 2013 by
edited Apr 11 by meena.p