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Question: Find the integration of the following, \(\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}} \) ?...

Find the integration of the following, ∫dxx(x2+1)\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}} ?

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Solution

The accumulation of discrete data is referred to as integration. The integral is used to determine the functions that will characterise the area, displacement, and volume that results from a collection of little data that cannot be measured individually. We have to learn about 10 different forms of integration. All questions are asked on that type.

Complete answer:
We have given ∫dxx(x2+1)\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}}
We will first change the integral part in partial fraction to make integration easy,
So 1x(x2+1)\dfrac{1}{{x\left( {{x^2} + 1} \right)}} can be written in partial form as
⇒1x(x2+1)=ax+bx+cx2+1\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{a}{x} + \dfrac{{bx + c}}{{{x^2} + 1}} (1)
We will simplify the left-hand side of equation 1,
⇒1x(x2+1)=a(x2+1)+(x)bx+cx(x2+1)\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{{a\left( {{x^2} + 1} \right) + \left( x \right)bx + c}}{{x\left( {{x^2} + 1} \right)}}
We will cancel the denominator on both side
1=(a+b)x2+cx+a1 = (a + b){x^2} + cx + a
We have compared the above equation, and get that
a=1, c=0, b= -1
we will put these values in equation 1, and get
⇒1x(x2+1)=1x+−xx2+1\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{1}{x} + \dfrac{{ - x}}{{{x^2} + 1}}
We will now integrate both sides,⇒∫1(x2+1)=∫1x+−xx2+1d′m=x=log⁡x−12log⁡∣1+x2∣+c \Rightarrow \int {\dfrac{1}{{\left( {{x^2} + 1} \right)}}} = \int {\dfrac{1}{x} + \dfrac{{ - x}}{{{x^2} + 1}}} d'm = x = \log x - \dfrac{1}{2}\log \mid 1 + {x^2}\mid + c
⇒∫1x(x2+1)=∫1x+−xx2+1\Rightarrow \int {\dfrac{1}{{x\left( {{x^2} + 1} \right)}}} = \int {\dfrac{1}{x} + \dfrac{{ - x}}{{{x^2} + 1}}}
We know that integration of 1x\dfrac{1}{x}is log⁡x\log x
We will solve the second part by using the formula ∫f′(x)f(x)=log⁡(f(x))\int {\dfrac{{f'(x)}}{{f(x)}}} = \log \left( {f(x)} \right)
We assume x2+1{x^2} + 1 as m and we know that d′m=xd'm = x
So, the integration of −xx2+1\dfrac{{ - x}}{{{x^2} + 1}} is −12log⁡∣1+x2∣- \dfrac{1}{2}\log \mid 1 + {x^2}\mid
We will now integrate RHS part wise, we get
=log⁡x−12log⁡∣1+x2∣+c= \log x - \dfrac{1}{2}\log \mid 1 + {x^2}\mid + c
So, the integration of ∫dxx(x2+1)\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}} is =log⁡x−12log⁡∣1+x2∣+c = \log x - \dfrac{1}{2}\log \mid 1 + {x^2}\mid + c.

Note:
An integral that contains the upper and lower limits then it is a definite integral. Indefinite integrals are defined without upper and lower limits. Integration can be used to find areas, volumes, central points and many useful things. One more point to be made in mind is that differential and integral are just opposite things.