Question

Evaluate the integral.

\(\int_{\rm{1}}^{\rm{e}} {\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + x + 1}}}}{{\rm{x}}}} {\rm{dx}}\).

Short answer

The solution of given integral\(\int_1^e {\frac{{{x^2} + x + 1}}{x}} dx\) is \(\frac{{{e^2} + 2e - 1}}{2}\).

Step-by-step solution

Power rule of integration

Power rule of integration allows us to find the indefinite and definite integrals of a variety of functions like polynomials, functions involving roots and even rational functions.

\(\int_a^b {{x^n}} dx = \left. {\frac{{{x^{n + 1}}}}{{n + 1}}} \right|_a^b\)

Applying power rule of integration

Using evaluation theorem write given integral as-

\(\begin{aligned}{c}\int_1^e {\frac{{{x^2} + x + 1}}{x}} dx = \int_1^e x + 1 + \frac{1}{x}dx\\ = \int_1^e x dx + \int_1^e 1 dx + \int_1^e {\frac{1}{x}} dx\\ = \left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_1^e + \left. {(x)} \right|_1^e + \left. {(\ln (x))} \right|_1^e\\ = \frac{{{e^2} - 1}}{2} + e - 1 + \ln (e) - \ln (1)\\ = \frac{{{e^2} + 2e - 1}}{2}\end{aligned}\)

Therefore, the solution of given integral\(\int_1^e {\frac{{{x^2} + x + 1}}{x}} dx\) is \(\frac{{{e^2} + 2e - 1}}{2}\).