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Expert-verified Found in: Page 335 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # evaluate: $$\int {ln({x^2} - x + 2)} dx$$

Apply integration by parts technique

For rational function, ensure the numerator has lower degree compared to denominator

Apply the rule $$\int {\frac{{f'(x)}}{{f(x)}}dx = {\mathop{\rm lnf}\nolimits} (x) + c}$$

Logarithmic function one difficult to be manipulated by the integration.

So use of integration by parts is advisable to keeping logarithmic function as a finite function.

One the function is turned into a rational function it can be treated like techniques like partial function and application of standard formula.

See the step by step solution

## Step 1: integration by parts

\begin{aligned}{l}I = \int {\ln ({x^2} - x + 2)dx = } \ln ({x^2} - x + 2)\int {dx - \int {\left( {\frac{d}{{dx}}\left( {\ln ({x^2} - x + 2)} \right)\int {dx} } \right)} } dx\\ = x\ln ({x^2} - x + 2) - \int {\frac{{x(2x - 1)}}{{{x^2} - x + 2}}dx + c} \\ = x\ln ({x^2} - x + 2) - {I_1} + c\end{aligned}

## Step 2: rationalization of $${I_1}$$

\begin{aligned}{l}{I_1} = \int {\frac{{x(2x - 1)}}{{{x^2} - x + 2}}dx = \int {\frac{{2({x^2} - x + 2) + x - 4}}{{{x^2} - x + 2}}} } dx\\ &= 2x + \int {\frac{{x - 4}}{{{x^2} - x + 2}}dx = 2x + {I_2}} \end{aligned}

## Step 3: rationalization of $${I_2}$$

\begin{aligned}{l}{I_2} &= \int {\frac{{x - 4}}{{{x^2} - x + 2}}dx} &= \frac{1}{2}\int {\frac{{2x - 1 - 7}}{{{x^2} - x + 2}}dx} \\ &= \frac{1}{2}\int {\frac{{\frac{d}{{dx}}({x^2} - x + 2)}}{{{x^2} - x + 2}}dx - \frac{7}{2}} \int {\frac{{dx}}{{{{(x - \frac{1}{2})}^2} + \frac{7}{2}}}} \\ &= \frac{1}{2}\ln ({x^2} - x + 2) - \frac{7}{2}{I_3}\end{aligned}

## Step 4: Evaluation of  $${I_3}$$ by using  $$\int {\frac{{dx}}{{{x^2} + {a^2}}}} = \frac{1}{a}ta{n^{ - 1}}\left( {\frac{x}{a}} \right)$$

$${I_3} = \int {\frac{{dx}}{{{{(x - \frac{1}{2})}^2} + {{\left( {\frac{{\sqrt 7 }}{2}} \right)}^2}}}} = \frac{1}{{\frac{{\sqrt 7 }}{2}}}{\tan ^{ - 1}}(\frac{{x - \frac{1}{2}}}{{\frac{{\sqrt 7 }}{2}}})$$

So we can write,

\begin{aligned}{l}I &= x\ln ({x^2} - x + 2) - 2x - \frac{1}{2}\ln ({x^2} - x + 2) + \frac{7}{2}\frac{2}{{\sqrt 7 }}{\tan ^{ - 1}}(\frac{{x - \frac{1}{2}}}{{\frac{{\sqrt 7 }}{2}}}) + c\\I &= (x - \frac{1}{2})\ln ({x^2} - x + 2) - 2x + \sqrt 7 {\tan ^{ - 1}}(\frac{{2x - 1}}{{\sqrt 2 }}) + c\end{aligned}

Hence the solution of the given integral will be:

$$I = (x - \frac{1}{2})\ln ({x^2} - x + 2) - 2x + \sqrt 7 {\tan ^{ - 1}}(\frac{{2x - 1}}{{\sqrt 2 }}) + c$$. ### Want to see more solutions like these? 