Final answer to the problem
$-\ln\left(\sqrt{-15+\left(s+5\right)^2}\right)+C_1+\ln\left(s\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{8.8729833+s}{\sqrt{-15+\left(s+5\right)^2}}\right)$
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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: $\ln\left(s\right)$
$\ln\left(s\right)$
Explain this step further
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 method of trigonometric substitution using the substitution
$s=\sqrt{15}\sec\left(\theta \right)-5$
Intermediate steps
15
Now, in order to rewrite $d\theta$ in terms of $ds$, we need to find the derivative of $s$. We need to calculate $ds$, we can do that by deriving the equation above
$ds=\sqrt{15}\sec\left(\theta \right)\tan\left(\theta \right)d\theta$
Explain this step further
16
Substituting in the original integral, we get
$\ln\left(s\right)-\int\frac{\sqrt{15}\left(\sqrt{15}\sec\left(\theta \right)-5\right)\sec\left(\theta \right)\tan\left(\theta \right)}{-15+15\sec\left(\theta \right)^2}d\theta$
17
Factor the polynomial $-15+15\sec\left(\theta \right)^2$ by it's greatest common factor (GCF): $15$
$\ln\left(s\right)-\int\frac{\sqrt{15}\left(\sqrt{15}\sec\left(\theta \right)-5\right)\sec\left(\theta \right)\tan\left(\theta \right)}{15\left(-1+\sec\left(\theta \right)^2\right)}d\theta$
18
Apply the trigonometric identity: $\sec\left(\theta \right)^2-1$$=\tan\left(\theta \right)^2$, where $x=\theta $
$\ln\left(s\right)-\int\frac{\sqrt{15}\left(\sqrt{15}\sec\left(\theta \right)-5\right)\sec\left(\theta \right)\tan\left(\theta \right)}{15\tan\left(\theta \right)^2}d\theta$
19
Taking the constant ($\sqrt{15}$) out of the integral
$\ln\left(s\right)-\sqrt{15}\int\frac{\left(\sqrt{15}\sec\left(\theta \right)-5\right)\sec\left(\theta \right)\tan\left(\theta \right)}{15\tan\left(\theta \right)^2}d\theta$
Intermediate steps
20
Simplify the expression inside the integral
$\ln\left(s\right)-\sqrt{15}\int\frac{\left(\sqrt{15}\sec\left(\theta \right)-5\right)\csc\left(\theta \right)}{15}d\theta$
Explain this step further
Intermediate steps
21
The integral $-\sqrt{15}\int\frac{\left(\sqrt{15}\sec\left(\theta \right)-5\right)\csc\left(\theta \right)}{15}d\theta$ results in: $-\ln\left(\frac{\sqrt{-15+\left(s+5\right)^2}}{\sqrt{15}}\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{s+5}{\sqrt{-15+\left(s+5\right)^2}}+\frac{\sqrt{15}}{\sqrt{-15+\left(s+5\right)^2}}\right)$
$-\ln\left(\frac{\sqrt{-15+\left(s+5\right)^2}}{\sqrt{15}}\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{s+5}{\sqrt{-15+\left(s+5\right)^2}}+\frac{\sqrt{15}}{\sqrt{-15+\left(s+5\right)^2}}\right)$
Explain this step further
22
Gather the results of all integrals
$\ln\left(s\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{s+5}{\sqrt{-15+\left(s+5\right)^2}}+\frac{\sqrt{15}}{\sqrt{-15+\left(s+5\right)^2}}\right)-\ln\left(\frac{\sqrt{-15+\left(s+5\right)^2}}{\sqrt{15}}\right)$
23
The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors
$L.C.M.=\sqrt{-15+\left(s+5\right)^2}$
Intermediate steps
24
Combine and simplify all terms in the same fraction with common denominator $\sqrt{-15+\left(s+5\right)^2}$
$\ln\left(s\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{8.8729833+s}{\sqrt{-15+\left(s+5\right)^2}}\right)-\ln\left(\frac{\sqrt{-15+\left(s+5\right)^2}}{\sqrt{15}}\right)$
Explain this step further
25
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\ln\left(s\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{8.8729833+s}{\sqrt{-15+\left(s+5\right)^2}}\right)-\ln\left(\frac{\sqrt{-15+\left(s+5\right)^2}}{\sqrt{15}}\right)+C_0$
26
Simplify the expression by applying logarithm properties
$-\ln\left(\sqrt{-15+\left(s+5\right)^2}\right)+C_1+\ln\left(s\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{8.8729833+s}{\sqrt{-15+\left(s+5\right)^2}}\right)$
Final answer to the problem
$-\ln\left(\sqrt{-15+\left(s+5\right)^2}\right)+C_1+\ln\left(s\right)-\frac{5}{\sqrt{15}}\ln\left(\frac{8.8729833+s}{\sqrt{-15+\left(s+5\right)^2}}\right)$