1.) Evaluate the indefinite integral using substitution method. \int \dfrac{e^{x} +1}{(e^{x}...


1.) Evaluate the indefinite integral using substitution method.

{eq}\displaystyle \int \dfrac{e^{x} +1}{(e^{x} +x)^2} dx {/eq}

2.) Using the answer in (1), evaluate the definite integral.

{eq}\displaystyle \int_{0}^{1} \dfrac{e^{x} +1}{(e^{x} +x)^2} dx {/eq}


{eq}\displaystyle{\text{This question is from integration and we have to find out the integral of the function and we add constant C after integrating this}\\[10pt] \text{because this is indefinite integral. We will use substitution method to solve this question.}\\[10pt] \text{Rule to evaluate}\int f(x)\quad dx \quad\text{by the substitution}\\[10pt] \text{1. In the integrand put x=g(y), dx = g'(y) dy}\\[10pt] \text{2. Evaluate the resulting integral in y.}\\[10pt] \text{3. Express the result obtained in terms of x.}} {/eq}

{eq}\displaystyle{\text{DEFINITE INTEGRAL} \int_a^b f(x) \ dx = \begin{vmatrix}F(x) \end{vmatrix}_a^b = F(b) - F(a)\\} . {/eq}

Answer and Explanation:

{eq}\displaystyle{ \int \dfrac{e^{x} +1}{(e^{x} +x)^2} dx\hspace{50pt}(1)\\[10pt] \text{Substitute u = } e^x + x\\[10pt] \frac{du}{dx} = e^x +1\\[10pt] du = e^x +1 dx\\[10pt] \text{ (1) becomes,}\\[10pt] \int \frac{1}{u^2} \ du\\[10pt] \int u^{-2} \ du\\[10pt] = \frac{u^{-1}}{-1} + C\hspace{100pt}\left(\text{ By applying power rule}\quad\int x^n \ dx = \frac{x^{n + 1}}{n + 1} + C\right)\\[10pt] = \frac {-1}{u} + C\\[10pt] = \frac{-1}{e^x + x} +C \\} {/eq}

Now applying the limit from 0 to 1 in the answer of question (1) we get,

{eq}\displaystyle{\begin{vmatrix}\frac{-1}{e^x + x}\end{vmatrix}_0^1\\[10pt] \left(\frac{-1}{e^1 + 1} - \frac{(-1)}{e^0 + 0}\right)\\[10pt] \left(\frac{-1}{e +1} + 1\right)\\[10pt] 1 -\frac{1}{e +1}\\[10pt] \boxed{\frac{e}{e + 1}}\hspace{80pt}\text{By taking L.C.M}\\} {/eq}

Learn more about this topic:

How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5

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