Question

\[\int\frac{x^2}{x^2 + 7x + 10} dx\]

Answer 1

\[Let I = \int\left( \frac{x^2}{x^2 + 7x + 10} \right)dx\]
\[Now, \]
\[ x^2 + 7x + 10 {x^2}^1 \]
\[ x^2 + 7x + 10\]
\[ - - - \]
\[ - 7x - 10 \]
\[ \therefore \frac{x^2}{x^2 + 7x + 10} = 1 - \frac{\left( 7x + 10 \right)}{x^2 + 7x + 10}\]
\[ \Rightarrow \frac{x^2}{x^2 + 7x + 10} = 1 - \left( \frac{7x + 10}{x^2 + 2x + 5x + 10} \right)\]
\[\frac{x^2}{x^2 + 7x + 10} = 1 - \left( \frac{7x + 10}{x \left( x + 2 \right) + 5 \left( x + 2 \right)} \right)\]
\[\frac{x^2}{x^2 + 7x + 10} = 1 - \left[ \frac{7x + 10}{\left( x + 2 \right) \left( x + 5 \right)} \right] . . . . . \left( 1 \right)\]
\[Consider, \]
\[\frac{7x + 10}{\left( x + 2 \right) \left( x + 5 \right)} = \frac{A}{\left( x + 2 \right)} + \frac{B}{x + 5}\]
\[7x + 10 = A \left( x + 5 \right) + B \left( x + 2 \right)\]
\[let x + 5 = 0\]
\[x = - 5\]
\[ \Rightarrow 7 \left( - 5 \right) + 10 = A \times 0 + B \left( - 5 + 2 \right)\]
\[ - 25 = B \left( - 3 \right)\]
\[ \Rightarrow B = \frac{25}{3}\]
\[let x + 2 = 0\]
\[x = - 2\]
\[7 \left( - 2 \right) + 10 = A \left( - 2 + 5 \right)\]
\[ \Rightarrow - 4 = A \left( 3 \right)\]
\[ \Rightarrow A = - \frac{4}{3}\]
\[\frac{7x + 10}{\left( x + 2 \right) \left( x + 5 \right)} = \frac{- 4}{3 \left( x + 2 \right)} + \frac{25}{3 \left( x + 5 \right)} . . . . . \left( 2 \right)\]
\[from \left( 1 \right) and \left( 2 \right)\]
\[\frac{x^2}{x^2 + 7x + 10} = 1 + \frac{4}{3 \left( x + 2 \right)} - \frac{25}{3 \left( x + 5 \right)}\]
\[ \Rightarrow \int\frac{x^2 dx}{x^2 + 7x + 10} = \int dx + \frac{4}{3}\int\frac{dx}{x + 2} - \frac{25}{3}\int\frac{dx}{x + 5}\]
\[ = x + \frac{4}{3} \log \left| x + 2 \right| - \frac{25}{3} \log \left| x + 5 \right| + C\]