 eiraszero11cu

2021-12-27

Evaluate the integral.
$\int \frac{{x}^{2}-5x+16}{\left(2x+1\right){\left(x-2\right)}^{2}}dx$ Toni Scott

Step 1
The integral given is, $\int \frac{{x}^{2}-5x+16}{\left(2x+1\right){\left(x-2\right)}^{2}}dx$.
After partial fraction decomposition,
$\int \frac{{x}^{2}-5x+16}{\left(2x+1\right){\left(x-2\right)}^{2}}dx=\int \frac{3}{2x+1}dx-\int \frac{1}{x-2}dx+\int \frac{2}{{\left(x-2\right)}^{2}}dx$.
Solve first part of integral ,
Substitute $u⇒2x+1,du=2dx$.
$\int \frac{3}{2x+1}dx=\frac{3}{2}\int \frac{du}{u}$
$=\frac{3}{2}\mathrm{ln}u+{c}_{1}$.
Step 2
After replace u with 2x+1,
$\int \frac{3}{2x+1}dx=\frac{3}{2}\mathrm{ln}\left(2x+1\right)+{c}_{1}$...(1)
Solving the second part of integral,
Substitute u\Rightarrow x-2, du=dx.
$\int \frac{dx}{x-2}=\int \frac{du}{u}$
$=\mathrm{ln}u+{c}_{2}$.
Replace u with x-2,
$\int \frac{dx}{x-2}=\mathrm{ln}\left(x-2\right)+{c}_{2}$...(2)
Step 3
Solving the third part of the integral,
Substitute u $⇒$ x-2, du=dx.
$\int \frac{dx}{{\left(x-2\right)}^{2}}=\int \frac{du}{{u}^{2}}$
$=\frac{-1}{u}+{c}_{3}$.
Replace u with x-2,
$\int \frac{dx}{{\left(x-2\right)}^{2}}=\frac{-1}{x-2}+{c}_{3}$...(3)
Step 4
The final solution of the integration becomes, adding equation 1 , 2 and 3,
$\int \frac{{x}^{2}-5x+16}{\left(2x+1\right){\left(x-2\right)}^{2}}dx=\int \frac{3}{2x+1}dx-\int \frac{1}{x-2}dx+\int \frac{2}{{\left(x-2\right)}^{2}}dx$. Debbie Moore

$\int \frac{{x}^{2}-5x+16}{\left(2x+1\right)×{\left(x-2\right)}^{2}}dx$
$\int \frac{3}{2x+1}-\frac{1}{x-2}+\frac{2}{{\left(x-2\right)}^{2}}dx$
$\int \frac{3}{2x+1}dx-\int \frac{1}{x-2}dx+\int \frac{2}{{\left(x-2\right)}^{2}}dx$
$\frac{3}{2}×\mathrm{ln}\left(|2x+1|\right)-\mathrm{ln}\left(|x-2|\right)-\frac{2}{x-2}$
Solution
$\frac{3}{2}×\mathrm{ln}\left(|2x+1|\right)-\mathrm{ln}\left(|x-2|\right)-\frac{2}{x-2}+C$ karton

$\begin{array}{}\int \frac{{x}^{2}-5x+16}{\left(x-2{\right)}^{2}\left(2x+1\right)}dx\\ =\int \left(\frac{3}{2x+1}-\frac{1}{x-2}+\frac{2}{\left(x-2{\right)}^{2}}\right)dx\\ =3\int \frac{1}{2x+1}dx-\int \frac{1}{x-2}dx+2\int \frac{1}{\left(x-2{\right)}^{2}}dx\\ \int \frac{1}{2x+1}dx\\ =\frac{1}{2}\int \frac{1}{u}du\\ \int \frac{1}{u}du\\ =\mathrm{ln}\left(u\right)\\ \frac{1}{2}\int \frac{1}{u}du\\ =\frac{\mathrm{ln}\left(u\right)}{2}\\ =\frac{\mathrm{ln}\left(2x+1\right)}{2}\\ \int \frac{1}{x-2}dx\\ =\int \frac{1}{u}du\\ =\mathrm{ln}\left(u\right)\\ =\mathrm{ln}\left(x-2\right)\\ \int \frac{1}{\left(x-2{\right)}^{2}}dx\\ =\int \frac{1}{{u}^{2}}du\\ =-\frac{1}{u}\\ =-\frac{1}{x-2}\\ 3\int \frac{1}{2x+1}dx-\int \frac{1}{x-2}dx+2\int \frac{1}{\left(x-2{\right)}^{2}}dx\\ =\frac{3\mathrm{ln}\left(2x+1\right)}{2}-\frac{2}{x-2}-\mathrm{ln}\left(x-2\right)\\ \int \frac{{x}^{2}-5x+16}{\left(x-2{\right)}^{2}\left(2x+1\right)}dx\\ =\frac{3\mathrm{ln}\left(|2x+1|\right)}{2}-\frac{2}{x-2}-\mathrm{ln}\left(|x-2|\right)+C\end{array}$