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Here, we show you a step-by-step solved example of integrals by partial fraction expansion. This solution was automatically generated by our smart calculator:
β«x(x+1)1βdx
ο Intermediate steps
Rewrite the fraction x(x+1)1β in 2 simpler fractions using partial fraction decomposition
x(x+1)1β=xAβ+x+1Bβ
Find the values for the unknown coefficients: A,B. The first step is to multiply both sides of the equation from the previous step by x(x+1)
1=x(x+1)(xAβ+x+1Bβ)
Multiplying polynomials
1=xx(x+1)Aβ+x+1x(x+1)Bβ
Simplifying
1=(x+1)A+xB
Assigning values to x we obtain the following system of equations
1=A1=βBβ(x=0)(x=β1)β
Proceed to solve the system of linear equations
1A0Aβ+ββ0B1Bβ=1=1β
Rewrite as a coefficient matrix
(10β0β1β11β)
Reducing the original matrix to a identity matrix using Gaussian Elimination
(10β01β1β1β)
The integral of x(x+1)1β in decomposed fractions equals
x1β+x+1β1β
2
Rewrite the fraction x(x+1)1β in 2 simpler fractions using partial fraction decomposition
x1β+x+1β1β
3
Expand the integral β«(x1β+x+1β1β)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately
β«x1βdx+β«x+1β1βdx
4
We can solve the integral β«x+1β1βdx 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 x+1 it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part
u=x+1
ο Intermediate steps
Differentiate both sides of the equation u=x+1
du=dxdβ(x+1)
Find the derivative
dxdβ(x+1)
The derivative of a sum of two or more functions is the sum of the derivatives of each function
dxdβ(x)+dxdβ(1)
The derivative of the constant function (1) is equal to zero
dxdβ(x)
The derivative of the linear function is equal to 1
1
5
Now, in order to rewrite dx in terms of du, we need to find the derivative of u. We need to calculate du, we can do that by finding the derivative of the equation above
du=dx
6
Substituting u and dx in the integral and simplify
β«x1βdx+β«uβ1βdu
ο Intermediate steps
The integral of the inverse of the lineal function is given by the following formula, β«x1βdx=ln(x)
lnβ£xβ£
7
The integral β«x1βdx results in: ln(x)
ln(x)
ο Intermediate steps
The integral of the inverse of the lineal function is given by the following formula, β«x1βdx=ln(x)
βlnβ£uβ£
Replace u with the value that we assigned to it in the beginning: x+1
βlnβ£x+1β£
8
The integral β«uβ1βdu results in: βln(x+1)
βln(x+1)
9
Gather the results of all integrals
lnβ£xβ£βlnβ£x+1β£
10
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C
lnβ£xβ£βlnβ£x+1β£+C0β
ξ Final answer to the exercise
lnβ£xβ£βlnβ£x+1β£+C0βξ
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