Question

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

Answer 1

Hint:We need to integrate the given fraction. For that, first of all, we see whether we can factorise the denominator and write it in terms of product of two or more terms. If we are able to do so and cancel out some terms from the numerator and denominator, we will obtain a simple expression which can be integrated using some integration formulas. If the expression cannot be factored, then we apply the By Parts formula.

Complete step by step answer:
We need to integrate \[\dfrac{{x + 1}}{{{x^2} + 2x + 1}}\]i.e. we need to find \[\int {\dfrac{{x + 1}}{{{x^2} + 2x + 1}}} dx\]. Let us consider
\[I = \int {\dfrac{{x + 1}}{{{x^2} + 2x + 1}}} dx\]
Now, let us see whether we can factorise the denominator or not.
By splitting the middle term in the denominator, we get
\[ \Rightarrow I = \int {\dfrac{{x + 1}}{{{x^2} + x + x + 1}}} dx\]

Now, clubbing first two terms and last two terms together in the denominator, we have
\[ \Rightarrow I = \int {\dfrac{{x + 1}}{{\left( {{x^2} + x} \right) + \left( {x + 1} \right)}}} dx\]
Taking out \[x\] common from the first two terms in the denominator, we get
\[ \Rightarrow I = \int {\dfrac{{x + 1}}{{\left( {x\left( {x + 1} \right)} \right) + 1\left( {x + 1} \right)}}} dx\]
Using the property \[\left( a \right) = 1\left( a \right)\] in the denominator, we get
\[ \Rightarrow I = \int {\dfrac{{x + 1}}{{\left( {x\left( {x + 1} \right)} \right) + \left( {1\left( {x + 1} \right)} \right)}}} dx\]

Now, taking out \[\left( {x + 1} \right)\] common from the denominator, we get
\[ \Rightarrow I = \int {\dfrac{{x + 1}}{{\left( {x + 1} \right)\left( {x + 1} \right)}}} dx\]
Now, we see that \[\left( {x + 1} \right)\]can be cancelled out from the numerator as well as the denominator.Cancelling out \[\left( {x + 1} \right)\] from the numerator as well as denominator, we get
\[ \Rightarrow I = \int {\dfrac{1}{{\left( {x + 1} \right)}}} dx\]
Now, Using the formula \[\int {\dfrac{1}{{\left( {x + a} \right)}}} dx = \ln |x + a| + c\], we get
\[ \therefore I = \int {\dfrac{1}{{\left( {x + 1} \right)}}} dx = \ln |x + 1| + c\], where \[c\] is a constant term.

Hence, we get, Integration of \[\dfrac{{x + 1}}{{{x^2} + 2x + 1}}\] is \[\ln |x + 1| + c\], where \[c\] is a constant term.

Note:We see that the denominator is of the form \[{a^2} + 2ab + {b^2}\] with \[a = x\] and \[b = 1\]. We know, \[{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}\]. So, instead of splitting the middle term formula, we can directly reduce it to this formula and then solve. We usually try to apply By Parts Formula when we see such a type of expression but we need to keep in mind that, we can apply that when we are not able to factorise the denominator or we have to use a product of two factors in the denominator.