Question

\({\bf{1 - 40}}\)- Evaluate the integral.

\(\int_1^2 {\frac{{{{(x + 1)}^2}}}{x}} dx\)

Short answer

\(\int_1^2 {\frac{{{{(x + 1)}^2}}}{x}} dx = \frac{7}{2} + \ln 2\)

Step-by-step solution

Definition

Integrationis a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation.

Simplify integrand

Consider the definite integral\(\int_1^2 {\frac{{{{(x + 1)}^2}}}{x}} dx\)

Using\({(a + b)^2} = {a^2} + 2ab + {b^2}\)we have:

\(\begin{aligned}{c}\int_1^2 {\frac{{{{(x + 1)}^2}}}{x}} dx = \int_1^2 {\frac{{{x^2} + 2x + 1}}{x}} dx\\ = \int_1^2 {\left( {x + 2 + \frac{1}{x}} \right)} dx\end{aligned}\)

Evaluate integral

Integrating we have:

\(\begin{aligned}{c}\int_1^2 {\frac{{{{(x + 1)}^2}}}{x}} dx = \left( {\frac{{{x^2}}}{2} + 2x + \ln |x|} \right)_1^2\\ = \left( {\frac{{{{(2)}^2}}}{2} + 2(2) + \ln 2} \right) - \left( {\frac{{{{(1)}^2}}}{2} + 2(1) + \ln 1} \right)\\ = (2 + 4 + \ln 2) - \left( {\frac{1}{2} + 2 + \ln 1} \right)\\ = \frac{7}{2} + \ln 2 - \ln 1\\ = \frac{7}{2} + \ln \left( {\frac{2}{1}} \right)\,\,\,\,\,\,\,{\rm{Since, }}\ln a - \ln b = \ln \left( {\frac{a}{b}} \right)\\ = \frac{7}{2} + \ln 2\end{aligned}\)

Therefore, the value of the given integral is\(\frac{7}{2} + \ln 2\).