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∫1x(x+1) dx
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1

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:

∫1x(x+1)dx\int\frac{1}{x\left(x+1\right)}dx

Rewrite the fraction 1x(x+1)\frac{1}{x\left(x+1\right)} in 22 simpler fractions using partial fraction decomposition

1x(x+1)=Ax+Bx+1\frac{1}{x\left(x+1\right)}=\frac{A}{x}+\frac{B}{x+1}

Find the values for the unknown coefficients: A,BA, B. The first step is to multiply both sides of the equation from the previous step by x(x+1)x\left(x+1\right)

1=x(x+1)(Ax+Bx+1)1=x\left(x+1\right)\left(\frac{A}{x}+\frac{B}{x+1}\right)

Multiplying polynomials

1=x(x+1)Ax+x(x+1)Bx+11=\frac{x\left(x+1\right)A}{x}+\frac{x\left(x+1\right)B}{x+1}

Simplifying

1=(x+1)A+xB1=\left(x+1\right)A+xB

Assigning values to xx we obtain the following system of equations

1=Aβ€…β€…β€…β€…β€…β€…β€…(x=0)1=βˆ’Bβ€…β€…β€…β€…β€…β€…β€…(x=βˆ’1)\begin{matrix}1=A&\:\:\:\:\:\:\:(x=0) \\ 1=-B&\:\:\:\:\:\:\:(x=-1)\end{matrix}

Proceed to solve the system of linear equations

1A+0B=10Aβˆ’1B=1\begin{matrix}1A & + & 0B & =1 \\ 0A & - & 1B & =1\end{matrix}

Rewrite as a coefficient matrix

(1010βˆ’11)\left(\begin{matrix}1 & 0 & 1 \\ 0 & -1 & 1\end{matrix}\right)

Reducing the original matrix to a identity matrix using Gaussian Elimination

(10101βˆ’1)\left(\begin{matrix}1 & 0 & 1 \\ 0 & 1 & -1\end{matrix}\right)

The integral of 1x(x+1)\frac{1}{x\left(x+1\right)} in decomposed fractions equals

1x+βˆ’1x+1\frac{1}{x}+\frac{-1}{x+1}
2

Rewrite the fraction 1x(x+1)\frac{1}{x\left(x+1\right)} in 22 simpler fractions using partial fraction decomposition

1x+βˆ’1x+1\frac{1}{x}+\frac{-1}{x+1}
3

Expand the integral ∫(1x+βˆ’1x+1)dx\int\left(\frac{1}{x}+\frac{-1}{x+1}\right)dx into 22 integrals using the sum rule for integrals, to then solve each integral separately

∫1xdx+βˆ«βˆ’1x+1dx\int\frac{1}{x}dx+\int\frac{-1}{x+1}dx
4

We can solve the integral βˆ«βˆ’1x+1dx\int\frac{-1}{x+1}dx 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 uu), which when substituted makes the integral easier. We see that x+1x+1 it's a good candidate for substitution. Let's define a variable uu and assign it to the choosen part

u=x+1u=x+1

Differentiate both sides of the equation u=x+1u=x+1

du=ddx(x+1)du=\frac{d}{dx}\left(x+1\right)

Find the derivative

ddx(x+1)\frac{d}{dx}\left(x+1\right)

The derivative of a sum of two or more functions is the sum of the derivatives of each function

ddx(x)+ddx(1)\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)

The derivative of the constant function (11) is equal to zero

ddx(x)\frac{d}{dx}\left(x\right)

The derivative of the linear function is equal to 11

11
5

Now, in order to rewrite dxdx in terms of dudu, we need to find the derivative of uu. We need to calculate dudu, we can do that by finding the derivative of the equation above

du=dxdu=dx
6

Substituting uu and dxdx in the integral and simplify

∫1xdx+βˆ«βˆ’1udu\int\frac{1}{x}dx+\int\frac{-1}{u}du

The integral of the inverse of the lineal function is given by the following formula, ∫1xdx=ln⁑(x)\displaystyle\int\frac{1}{x}dx=\ln(x)

ln⁑∣x∣\ln\left|x\right|
7

The integral ∫1xdx\int\frac{1}{x}dx results in: ln⁑(x)\ln\left(x\right)

ln⁑(x)\ln\left(x\right)

The integral of the inverse of the lineal function is given by the following formula, ∫1xdx=ln⁑(x)\displaystyle\int\frac{1}{x}dx=\ln(x)

βˆ’ln⁑∣u∣-\ln\left|u\right|

Replace uu with the value that we assigned to it in the beginning: x+1x+1

βˆ’ln⁑∣x+1∣-\ln\left|x+1\right|
8

The integral βˆ«βˆ’1udu\int\frac{-1}{u}du results in: βˆ’ln⁑(x+1)-\ln\left(x+1\right)

βˆ’ln⁑(x+1)-\ln\left(x+1\right)
9

Gather the results of all integrals

ln⁑∣xβˆ£βˆ’ln⁑∣x+1∣\ln\left|x\right|-\ln\left|x+1\right|
10

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

ln⁑∣xβˆ£βˆ’ln⁑∣x+1∣+C0\ln\left|x\right|-\ln\left|x+1\right|+C_0

ξ ƒ Final answer to the exercise

ln⁑∣xβˆ£βˆ’ln⁑∣x+1∣+C0\ln\left|x\right|-\ln\left|x+1\right|+C_0 ξ ƒ

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