use any method to evaluate the integral.\n int\frac{x^{3}dx}{x^{2}-49}\n int\frac{x^{3}dx}{x^{2}-49}=square…

use any method to evaluate the integral.\n int\frac{x^{3}dx}{x^{2}-49}\n int\frac{x^{3}dx}{x^{2}-49}=square (use parentheses to clearly denote the argument of each function)

use any method to evaluate the integral.\n int\frac{x^{3}dx}{x^{2}-49}\n int\frac{x^{3}dx}{x^{2}-49}=square (use parentheses to clearly denote the argument of each function)

Answer

Explanation:

Step1: Perform polynomial long - division

Since the degree of the numerator ($x^{3}$) is greater than the degree of the denominator ($x^{2}-49$), we divide $x^{3}$ by $x^{2}-49$. $x^{3}=(x)(x^{2}-49)+49x$. So, $\frac{x^{3}}{x^{2}-49}=x+\frac{49x}{x^{2}-49}$.

Step2: Integrate term - by - term

$\int\frac{x^{3}}{x^{2}-49}dx=\int xdx+\int\frac{49x}{x^{2}-49}dx$. The integral of $x$ is $\int xdx=\frac{x^{2}}{2}+C_1$. For $\int\frac{49x}{x^{2}-49}dx$, let $u = x^{2}-49$, then $du = 2xdx$ and $xdx=\frac{1}{2}du$. So, $\int\frac{49x}{x^{2}-49}dx=\frac{49}{2}\int\frac{du}{u}$.

Step3: Integrate $\frac{49}{2}\int\frac{du}{u}$

We know that $\int\frac{du}{u}=\ln|u|+C_2$. Substituting back $u = x^{2}-49$, we get $\frac{49}{2}\ln|x^{2}-49|+C_2$.

Step4: Combine the results

$\int\frac{x^{3}}{x^{2}-49}dx=\frac{x^{2}}{2}+\frac{49}{2}\ln|x^{2}-49|+C$, where $C = C_1 + C_2$.

Answer:

$\frac{x^{2}}{2}+\frac{49}{2}\ln|x^{2}-49|+C$