evaluate the indefinite integral given below.\n int\frac{- 5x^{4}-x^{2}}{-3x^{5}-x^{3}}dx=square \nprovide…

Explanation:

Step1: Simplify the integrand

First, simplify $\frac{- 5x^{4}-x^{2}}{-3x^{5}-x^{3}}$. Factor out common - factors: $\frac{-x^{2}(5x^{2} + 1)}{-x^{3}(3x^{2}+1)}=\frac{5x^{2}+1}{x(3x^{2}+1)}$. Use partial - fraction decomposition. Let $\frac{5x^{2}+1}{x(3x^{2}+1)}=\frac{A}{x}+\frac{Bx + C}{3x^{2}+1}$. Then $5x^{2}+1=A(3x^{2}+1)+(Bx + C)x=3Ax^{2}+A + Bx^{2}+Cx=(3A + B)x^{2}+Cx+A$. We have the system of equations: $\begin{cases}3A + B = 5\C = 0\A = 1\end{cases}$. Substitute $A = 1$ into $3A + B = 5$, we get $3\times1 + B = 5$, so $B = 2$ and $C = 0$. So $\frac{5x^{2}+1}{x(3x^{2}+1)}=\frac{1}{x}+\frac{2x}{3x^{2}+1}$.

Step2: Integrate term - by - term

The integral $\int\frac{- 5x^{4}-x^{2}}{-3x^{5}-x^{3}}dx=\int(\frac{1}{x}+\frac{2x}{3x^{2}+1})dx$. We know that $\int\frac{1}{x}dx=\ln|x|+C_1$ and for $\int\frac{2x}{3x^{2}+1}dx$, let $u = 3x^{2}+1$, then $du=6xdx$, and $\int\frac{2x}{3x^{2}+1}dx=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}\ln|u|+C_2=\frac{1}{3}\ln|3x^{2}+1|+C_2$.

Step3: Combine the results

$\int(\frac{1}{x}+\frac{2x}{3x^{2}+1})dx=\ln|x|+\frac{1}{3}\ln|3x^{2}+1|+C$, where $C = C_1 + C_2$.

Answer:

$\ln|x|+\frac{1}{3}\ln|3x^{2}+1|+C$