Solveeit Logo
Login

Question

Question: How do you integrate \( \int \dfrac{{{x^2} + 2x}}{{{x^2} + 2x + 1}}dx \) using substitution?...

How do you integrate ∫x2+2xx2+2x+1dx\int \dfrac{{{x^2} + 2x}}{{{x^2} + 2x + 1}}dx using substitution?

Explanation

Solution

To solve this question, first we will make the given integration expression simple to integrate further. Then assume one of them factors in the expression and we can easily proceed to the conclusion.

Complete step by step solution:
To make the given expression simple, first we will factorise the given expression:
∫(x2+2x)x2+2x+1dx =∫x(x+2)(x+1)2dx   \int \dfrac{{({x^2} + 2x)}}{{{x^2} + 2x + 1}}dx \\\ = \int \dfrac{{x(x + 2)}}{{{{(x + 1)}^2}}}dx \;
Now, let (x+1)(x + 1) be uu or u=x+1u = x + 1 .
Differentiate u=x+1u = x + 1 :

⇒dudx=1+0 ⇒du=dx   \Rightarrow \dfrac{{du}}{{dx}} = 1 + 0 \\\ \Rightarrow du = dx \;

∵u=x+1 ⇒u+1=x+2 ∴x=u−1   \because u = x + 1 \\\ \Rightarrow u + 1 = x + 2 \\\ \therefore x = u - 1 \;
So,
∫x(x+2)dx(x+1)2=∫(u+1)(u−1)duu2 =∫(u2−1)u2du =∫(1−1u2)du =u+1u =(x+1)+1x+1+C   \int \dfrac{{x(x + 2)dx}}{{{{(x + 1)}^2}}} = \int \dfrac{{(u + 1)(u - 1)du}}{{{u^2}}} \\\ = \int \dfrac{{({u^2} - 1)}}{{{u^2}}}du \\\ = \int (1 - \dfrac{1}{{{u^2}}})du \\\ = u + \dfrac{1}{u} \\\ = (x + 1) + \dfrac{1}{{x + 1}} + C \;
So, the correct answer is “ (x+1)+1x+1+C(x + 1) + \dfrac{1}{{x + 1}} + C ”.

Note : - Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.