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Find the integral $\int\frac{10s+10}{s\left(s^2+10s+10\right)}ds$

Step-by-step Solution

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Final answer to the problem

$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)+C_0$
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Step-by-step Solution

Specify the solving method

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

$10s+10=\frac{s\left(s^2+10s+10\right)A}{s}+\frac{s\left(s^2+10s+10\right)\left(Bs+C\right)}{s^2+10s+10}$
4

Simplifying

$10s+10=\left(s^2+10s+10\right)A+s\left(Bs+C\right)$
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

$\int\left(\frac{1}{s}+\frac{-s}{s^2+10s+10}\right)ds$
10

Simplify the expression inside the integral

$\int\frac{1}{s}ds-\int\frac{s}{s^2+10s+10}ds$
11

The integral $\int\frac{1}{s}ds$ results in: $\ln\left(s\right)$

$\ln\left(s\right)$
12

Gather the results of all integrals

$\ln\left(s\right)-\int\frac{s}{s^2+10s+10}ds$
13

Rewrite the expression $\frac{s}{s^2+10s+10}$ inside the integral in factored form

$\ln\left(s\right)-\int\frac{s}{-15+\left(s+5\right)^2}ds$
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

$u=s+5$
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

$du=ds$
16

Rewriting $s$ in terms of $u$

$s=u-5$
17

Substituting $u$, $ds$ and $s$ in the integral and simplify

$\ln\left(s\right)-\int\frac{u-5}{-15+u^2}du$
18

The integral $-\int\frac{u-5}{-15+u^2}du$ results in: $-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(s+5-\sqrt{15}\right)$

$-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(s+5-\sqrt{15}\right)$
19

Gather the results of all integrals

$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(s+5-\sqrt{15}\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)$
20

Subtract the values $5$ and $-\sqrt{15}$

$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(\sqrt{15}+s+5\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)$
21

Add the values $\sqrt{15}$ and $5$

$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)$
22

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)+C_0$

Final answer to the problem

$\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)+C_0$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of ((10s+10)/s(s^2+10s))ds using partial fractionsSolve integral of ((10s+10)/s(s^2+10s))ds using basic integralsSolve integral of ((10s+10)/s(s^2+10s))ds using integration by partsSolve integral of ((10s+10)/s(s^2+10s))ds using trigonometric substitution

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Function Plot

Plotting: $\ln\left(s\right)-\frac{5}{2\sqrt{15}}\ln\left(8.8729833+s\right)+\frac{5}{2\sqrt{15}}\ln\left(1.1270167+s\right)-\frac{1}{2}\ln\left(-15+\left(s+5\right)^2\right)+C_0$

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0
a
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g
m
n
u
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x
y
z
.
(◻)
+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Integrals by Partial Fraction Expansion

The partial fraction decomposition or partial fraction expansion of a rational function is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

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