Final answer to the problem
$\ln\left(s\right)-1.1454972\ln\left(8.8729833+s\right)+\frac{55}{378}\ln\left(1.1270167+s\right)+C_0$
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Step-by-step Solution
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1
Rewrite the fraction $\frac{10s+10}{s\left(s^2+10s+10\right)}$ in $2$ simpler fractions using partial fraction decomposition
$\frac{10s+10}{s\left(s^2+10s+10\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+10s+10}$
2
Find the values for the unknown coefficients: $A, B, C$. The first step is to multiply both sides of the equation from the previous step by $s\left(s^2+10s+10\right)$
$10s+10=s\left(s^2+10s+10\right)\left(\frac{A}{s}+\frac{Bs+C}{s^2+10s+10}\right)$
3
Multiplying polynomials
$10s+10=\frac{s\left(s^2+10s+10\right)A}{s}+\frac{s\left(s^2+10s+10\right)\left(Bs+C\right)}{s^2+10s+10}$
$10s+10=\left(s^2+10s+10\right)A+s\left(Bs+C\right)$
5
Assigning values to $s$ we obtain the following system of equations
$\begin{matrix}10=10A&\:\:\:\:\:\:\:(s=0) \\ -90=10A+100B-10C&\:\:\:\:\:\:\:(s=-10) \\ 110=210A+100B+10C&\:\:\:\:\:\:\:(s=10)\end{matrix}$
6
Proceed to solve the system of linear equations
$\begin{matrix}10A & + & 0B & + & 0C & =10 \\ 10A & + & 100B & - & 10C & =-90 \\ 210A & + & 100B & + & 10C & =110\end{matrix}$
7
Rewrite as a coefficient matrix
$\left(\begin{matrix}10 & 0 & 0 & 10 \\ 10 & 100 & -10 & -90 \\ 210 & 100 & 10 & 110\end{matrix}\right)$
8
Reducing the original matrix to a identity matrix using Gaussian Elimination
$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0\end{matrix}\right)$
9
The integral of $\frac{10s+10}{s\left(s^2+10s+10\right)}$ in decomposed fraction equals
$\int\left(\frac{1}{s}+\frac{-s}{s^2+10s+10}\right)ds$
Intermediate steps
10
Simplify the expression inside the integral
$\int\frac{1}{s}ds-\int\frac{s}{s^2+10s+10}ds$
Explain this step further
Intermediate steps
11
The integral $\int\frac{1}{s}ds$ results in: $\ln\left(s\right)$
$\ln\left(s\right)$
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12
Gather the results of all integrals
$\ln\left(s\right)-\int\frac{s}{s^2+10s+10}ds$
Intermediate steps
13
Rewrite the expression $\frac{s}{s^2+10s+10}$ inside the integral in factored form
$\ln\left(s\right)-\int\frac{s}{-15+\left(s+5\right)^2}ds$
Explain this step further
14
We can solve the integral $\int\frac{s}{-15+\left(s+5\right)^2}ds$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $s+5$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
$u=s+5$
Intermediate steps
15
Now, in order to rewrite $ds$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
$du=ds$
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Intermediate steps
16
Rewriting $s$ in terms of $u$
$s=u-5$
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Intermediate steps
17
Substituting $u$, $ds$ and $s$ in the integral and simplify
$\ln\left(s\right)-\int\frac{u-5}{-15+u^2}du$
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Intermediate steps
18
The integral $-\int\frac{u-5}{-15+u^2}du$ results in: $-\frac{1}{2}\ln\left(\sqrt{15}+s+5\right)-\frac{1}{2}\ln\left(s+5-\sqrt{15}\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(s+5-\sqrt{15}\right)$
$-\frac{1}{2}\ln\left(\sqrt{15}+s+5\right)-\frac{1}{2}\ln\left(s+5-\sqrt{15}\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(s+5-\sqrt{15}\right)$
Explain this step further
19
Gather the results of all integrals
$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(s+5-\sqrt{15}\right)-\frac{1}{2}\ln\left(\sqrt{15}+s+5\right)-\frac{1}{2}\ln\left(s+5-\sqrt{15}\right)$
20
Subtract the values $5$ and $-\sqrt{15}$
$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(\sqrt{15}+s+5\right)-\frac{1}{2}\ln\left(s+5-\sqrt{15}\right)$
21
Add the values $\sqrt{15}$ and $5$
$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(\sqrt{15}+s+5\right)-\frac{1}{2}\ln\left(s+5-\sqrt{15}\right)$
22
Subtract the values $5$ and $-\sqrt{15}$
$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(\sqrt{15}+s+5\right)-\frac{1}{2}\ln\left(1.1270167+s\right)$
23
Add the values $\sqrt{15}$ and $5$
$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(8.8729833+s\right)-\frac{1}{2}\ln\left(1.1270167+s\right)$
24
Combining like terms $-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)$ and $-\frac{1}{2}\ln\left(8.8729833+s\right)$
$\ln\left(s\right)-1.1454972\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(1.1270167+s\right)$
25
Combining like terms $\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)$ and $-\frac{1}{2}\ln\left(1.1270167+s\right)$
$\ln\left(s\right)-1.1454972\ln\left(8.8729833+s\right)+\frac{55}{378}\ln\left(1.1270167+s\right)$
26
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\ln\left(s\right)-1.1454972\ln\left(8.8729833+s\right)+\frac{55}{378}\ln\left(1.1270167+s\right)+C_0$
Final answer to the problem
$\ln\left(s\right)-1.1454972\ln\left(8.8729833+s\right)+\frac{55}{378}\ln\left(1.1270167+s\right)+C_0$