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$\Large \int\frac{x^3}{x+1}$ is equal to\begin{array}{1 1}(A)\;x+\frac{x^2}{2}+\frac{x^3}{3}-log\mid1-x\mid+C & (B)\;x+\frac{x^2}{2}-\frac{x^3}{3}-log\mid1-x\mid+C\\(C)\;x-\frac{x^2}{2}-\frac{x^3}{3}-log\mid1+x \mid+C & (D)\;x-\frac{x^2}{2}+\frac{x^3}{3}-log\mid1+x \mid+C\end{array}

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  • If the degree of the numerator in a rational expression is greater than the denominator,then it is an improper fraction.
  • $\int x^ndx=\large\frac{x^{n+1}}{n+1}$+c.
  • $\int \large\frac{dx}{x+a}=\normalsize \log\mid x+a\mid +c.$
Step 1:
$\large\frac{x^3}{x+1}$ is an improper rational expression,to make proper let us divide
$\int\large \frac{x^3}{x+1}$=$\int (x^2-x+1)dx-\int \large\frac{1}{x+1}\normalsize dx.$
Step 2:
On integrating we get,
$\large\frac{x^3}{3}-\frac{x^2}{2}$+$x-\log\mid x+1\mid+c.$
Correct answer is D.
answered Apr 24, 2013 by sreemathi.v

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