Solveeit Logo
Login

Question

Question: How do you integrate\[\dfrac{1}{{{x}^{2}}+x}\]?...

How do you integrate1x2+x\dfrac{1}{{{x}^{2}}+x}?

Explanation

Solution

This question is from the topic of integration of chapter calculus. In this question, we are going to use this formula: ∫1xdx=ln⁡x\int{\dfrac{1}{x}dx}=\ln x. During solving this question, first we will make the term 1x2+x\dfrac{1}{{{x}^{2}}+x} in simplest form, we can do the integration easily. After converting the term in simplest form, we will use the formula that we have seen in above. After that, we will arrange the term and find the integration.

Complete step-by-step answer:
Let us solve this question.
In this question, we have asked to integrate the term1x2+x\dfrac{1}{{{x}^{2}}+x}.
Before integrating the term1x2+x\dfrac{1}{{{x}^{2}}+x}, let us recall a formula that we are going to use here for the integration. The formula for integration is:
∫1xdx=ln⁡x+C\int{\dfrac{1}{x}dx}=\ln x+C, where C is the constant
Let, I=∫(1x2+x)dxI=\int{\left( \dfrac{1}{{{x}^{2}}+x} \right)dx}
Now, we write the term 1x2+x\dfrac{1}{{{x}^{2}}+x} in simple form to make the integration easier
We can write
⇒I=∫(1x2+x)dx=∫(1x(x+1))dx\Rightarrow I=\int{\left( \dfrac{1}{{{x}^{2}}+x} \right)dx}=\int{\left( \dfrac{1}{x\left( x+1 \right)} \right)dx}
The above equation can also be written as
⇒I=∫(x+1−xx(x+1))dx\Rightarrow I=\int{\left( \dfrac{x+1-x}{x\left( x+1 \right)} \right)dx}
We can write the above equation as
⇒I=∫(x+1x(x+1)−xx(x+1))dx\Rightarrow I=\int{\left( \dfrac{x+1}{x\left( x+1 \right)}-\dfrac{x}{x\left( x+1 \right)} \right)dx}
The above equation can also be written as
⇒I=∫(1x−1x+1)dx\Rightarrow I=\int{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)dx}
The above equation can also be written as
⇒I=∫1xdx−∫1x+1dx\Rightarrow I=\int{\dfrac{1}{x}dx}-\int{\dfrac{1}{x+1}dx}
Using the formula of integration ∫1xdx=ln⁡x\int{\dfrac{1}{x}dx}=\ln x here, we get
⇒I=ln⁡x−ln⁡(x+1)+C\Rightarrow I=\ln x-\ln (x+1)+C
As we know that, ln⁡ab=ln⁡a−ln⁡b\ln \dfrac{a}{b}=\ln a-\ln b. So, using this in the above equation, we get
⇒I=ln⁡xx+1+C\Rightarrow I=\ln \dfrac{x}{x+1}+C
Hence, we get that integration of the term 1x2+x\dfrac{1}{{{x}^{2}}+x} is ln⁡xx+1+C\ln \dfrac{x}{x+1}+C

Note: For solving this type of question, we should have a better knowledge in integration. And, also we should know the formulas of integration. So that we can solve this type of question easily. Remember the formula for integration like: ∫1xdx=ln⁡x+C\int{\dfrac{1}{x}dx}=\ln x+C for solving this type of question easily. And, also don’t forget the formula of logarithms like ln⁡ab=ln⁡a−ln⁡b\ln \dfrac{a}{b}=\ln a-\ln b.
We have an alternate method to solve this question.
We can solve this question using a partial fraction method.
In this method, we will have to decompose the term 1x2+x\dfrac{1}{{{x}^{2}}+x}.
Let us decompose that term.
1x2+x=1x(x+1)=Ax+Bx+1\dfrac{1}{{{x}^{2}}+x}=\dfrac{1}{x\left( x+1 \right)}=\dfrac{A}{x}+\dfrac{B}{x+1}
The above equation can also be written as
⇒1=A(x(x+1))x+B(x(x+1))x+1\Rightarrow 1=\dfrac{A\left( x\left( x+1 \right) \right)}{x}+\dfrac{B\left( x\left( x+1 \right) \right)}{x+1}
We can write the above equation as
⇒1=A(x+1)+Bx\Rightarrow 1=A\left( x+1 \right)+Bx
We can write the above equation as
⇒1=A+Ax+Bx\Rightarrow 1=A+Ax+Bx
The above equation can also be written as
⇒1=A+(A+B)x\Rightarrow 1=A+\left( A+B \right)x
Now, comparing the coefficients of x and constant, we get
A=1 and A+B=0
So, we get that A=1 and B=-1
So, we can say that the term1x2+x\dfrac{1}{{{x}^{2}}+x} can also be written as 1x+(−1)x+1\dfrac{1}{x}+\dfrac{\left( -1 \right)}{x+1} which can also be written as 1x−1x+1\dfrac{1}{x}-\dfrac{1}{x+1}.
So, from here we have integrated the term in the above solution. Hence, one can take reference from there.