integrate. :- x/(x²+x-6)
Answers
Answer:
Step-by-step explanation:
First, we have
\[ \frac{x^2}{x^2 + x - 6} = 1 - \frac{x-6}{x^2 + x - 6} = 1 - \frac{x-6}{(x+3)(x-2)}. \]
Therefore,
\begin{align*} \int \frac{x^2}{x^2 + x - 6} &= \int \left( 1 - \frac{x-6}{(x+3)(x-2)} \right) \, dx \\ &= x - \int \frac{x-6}{(x+3)(x-2)} \, dx. \end{align*}
We use partial fractions to evaluate the integral on the right. To that end, we write
\[ \frac{x-6}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}. \]
This gives us the equation
\[ A(x-2) + B(x+3) = x-6. \]
Evaluating at x = -3 and x = 2 we then have
\begin{align*} -5A &= -9 & \implies \qquad A &= \frac{9}{5} \\ 5B &= -4 & \implies \qquad B &= -\frac{4}{5}. \end{align*}
Therefore,
\begin{align*} \int \frac{x^2}{x^2 + x - 6} \, dx &= x - \int \frac{x-6}{(x+3)(x-2)} \, dx \\ &= x - \frac{9}{5} \int \frac{1}{x+3} \, dx + \frac{4}{5} \int \frac{1}{x-2} \, dx \\ &= x - \frac{9}{5} \log |x+3| + \frac{4}{5} \log |x-2| + C \\ &= x + \frac{4}{5} \log |x-2| - \frac{9}{5} \log |x+3| + C. \end{align*}
Answer:
Refer to attachment ❤✨
