Question

How do you integrate $\ln ({x^2} - x + 2)dx$?

Answer 1

We are asked to integrate the function $\ln ({x^2} - x + 2)dx$, where $\ln$ represents the natural logarithm to the base $e$. Firstly, we make use of methods of integration by parts. The method is given by $\int {udv = uv - \int {vdu} }$. We take here $u = \ln ({x^2} - x + 2)$ and $dv = 1$ to solve. Then we make use of some properties of integration to simplify the problem and obtain the solution.

Complete step by step solution:
In this question we need to integrate a function $\ln ({x^2} - x + 2)dx$.
Note that logarithm is an inverse of an exponential function. Here $\ln$represents the natural logarithm to the base $e$.
To solve the above problem we use the method of integration by parts.
The method is given by $\int {udv = uv - \int {vdu} }$ …… (1)
Take $u = \ln ({x^2} - x + 2)$
Differentiating u we obtain $du$.
$du = \dfrac{{2x - 1}}{{{x^2} - x + 2}}$
Take $dv = 1$.
Integrate $dv$ to obtain v.
$\int {dv = \int {1dx} }$
$\Rightarrow v = x$.
Now substituting in the equation (1), we get,
$\int {\ln ({x^2}} - x + 2) \cdot 1dx = \ln ({x^2} - x + 2)x - \int {x\dfrac{{2x - 1}}{{{x^2} - x + 2}}} dx$
$\Rightarrow \int {\ln ({x^2}} - x + 2)dx = x \cdot \ln ({x^2} - x + 2) - \int {\dfrac{{2{x^2} - x}}{{{x^2} - x + 2}}} dx$ …… (2)
Now we simplify the term $\dfrac{{2{x^2} - x}}{{{x^2} - x + 2}}$.
Express $- x$ as $- 2x + x$, we get,
$\Rightarrow \dfrac{{2{x^2} - 2x + x}}{{{x^2} - x + 2}}$
Now add and subtract by 4 in the numerator we get,
$\Rightarrow \dfrac{{2{x^2} - 2x + 4 + x - 4}}{{{x^2} - x + 2}}$
Now we split the above term as,
$\Rightarrow \dfrac{{2({x^2} - x + 2)}}{{{x^2} - x + 2}} + \dfrac{{x - 4}}{{{x^2} - x + 2}}$
$\Rightarrow 2 + \dfrac{{x - 4}}{{{x^2} - x + 2}}$
Now $\int {\dfrac{{2{x^2} - x}}{{{x^2} - x + 2}}dx} = \int {2 + } \dfrac{{x - 4}}{{{x^2} - x + 2}}dx$
$\Rightarrow \int {2dx + \int {\dfrac{{x - 4}}{{{x^2} - x + 2}}} } dx$ …… (3)
Now we simplify the term $\dfrac{{x - 4}}{{{x^2} - x + 2}}$
We try to convert it as $\dfrac{{f'(x)}}{{f(x)}}$.
$\dfrac{{x - 4}}{{{x^2} - x + 2}} = \dfrac{{\dfrac{1}{2}(2x - 8)}}{{{x^2} - x + 2}}$
$\Rightarrow \dfrac{{\dfrac{1}{2}(2x - 1) - \dfrac{7}{2}}}{{{x^2} - x + 2}}$
Now splitting the terms we get,
$\Rightarrow \dfrac{{\dfrac{1}{2}(2x - 1)}}{{{x^2} - x + 2}} - \dfrac{7}{2}\dfrac{1}{{{x^2} - x + 2}}$
Substituting this in the equation (3), we get,
$\Rightarrow \int {2dx + \int {\left( {\dfrac{{\dfrac{1}{2}(2x - 1)}}{{{x^2} - x + 2}} - \dfrac{7}{2}\dfrac{1}{{{x^2} - x + 2}}} \right)dx} }$
Splitting the integration in the second term we get,
$\Rightarrow \int {2dx + } \int {\dfrac{{\dfrac{1}{2}(2x - 1)}}{{{x^2} - x + 2}}} dx - \int {\dfrac{7}{2}\dfrac{1}{{{x^2} - x + 2}}dx}$
Now integrating we get,
$\Rightarrow 2x + \dfrac{1}{2}\ln ({x^2} - x + 2) - \dfrac{7}{2}\int {\dfrac{1}{{{x^2} - x + 2}}dx}$
Substituting in the equation (2) we get,
$\int {\ln ({x^2}} - x + 2)dx = x \cdot \ln ({x^2} - x + 2) - \int {\dfrac{{2{x^2} - x}}{{{x^2} - x + 2}}} dx$
$\Rightarrow x\ln ({x^2} - x + 2) - \left( {2x + \dfrac{1}{2}\ln ({x^2} - x + 2) - \dfrac{7}{2}\int {\dfrac{1}{{{x^2} - x + 2}}dx} } \right)$
Now combining the like terms we get,
$\Rightarrow \left( {x - \dfrac{1}{2}} \right)\ln ({x^2} - x + 2) - 2x + \dfrac{7}{2}\int {\dfrac{1}{{{x^2} - x + 2}}dx}$ …… (4)
Now we simplify the term $\int {\dfrac{1}{{{x^2} - x + 2}}dx}$.
We try to convert the denominator as factors using completing squares.
Consider the quadratic equation ${x^2} - x + 2$.
Add and subtract $\dfrac{1}{4}$ we get,
$\Rightarrow {x^2} - x + \dfrac{1}{4} + 2 - \dfrac{1}{4}$
$\Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4}$
$\therefore \int {\dfrac{1}{{{x^2} - x + 2}}dx} = \int {\dfrac{1}{{{{\left( {x - \dfrac{1}{2}} \right)}^2} + \dfrac{7}{4}}}} dx$ …… (5)
Take ${\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{7}{4}{\tan ^2}\theta$
$\Rightarrow {\left( {x - \dfrac{1}{2}} \right)^{}} = \dfrac{{\sqrt 7 }}{2}\tan \theta$ …… (6)
$\Rightarrow \theta = \arctan \dfrac{2}{{\sqrt 7 }}\left( {x - \dfrac{1}{2}} \right)$ …… (7)
Differentiating (6) we get,
$\Rightarrow dx = \dfrac{{\sqrt 7 }}{2}{\sec ^2}\theta d\theta$
Hence the equation (5) becomes,
$\Rightarrow \int {\dfrac{1}{{\dfrac{7}{4}{{\tan }^2}\theta + \dfrac{7}{4}}}} \cdot \dfrac{{\sqrt 7 }}{2}{\sec ^2}\theta d\theta$
$\Rightarrow \dfrac{{\sqrt 7 }}{2}\int {\dfrac{1}{{\dfrac{7}{4}({{\tan }^2}\theta + 1)}}} \cdot {\sec ^2}\theta d\theta$
We know the trigonometric identity $1 + {\tan ^2}\theta = {\sec ^2}\theta$.
$\Rightarrow \dfrac{{\sqrt 7 }}{2}\int {\dfrac{{{{\sec }^2}\theta }}{{\dfrac{7}{4}({{\sec }^2}\theta )}}d\theta }$
Cancelling the terms and taking out the constant term in the denominator out of integration we get,
$\Rightarrow \dfrac{{\sqrt 7 }}{2} \cdot \dfrac{2}{7}\int {d\theta }$
Note that $7 = \sqrt 7 \times \sqrt 7$
$\Rightarrow \dfrac{{\sqrt 7 }}{2} \cdot \dfrac{2}{{\sqrt 7 \times \sqrt 7 }}\int {d\theta }$
$\Rightarrow \dfrac{2}{{\sqrt 7 }}\int {d\theta }$
$\Rightarrow \dfrac{2}{{\sqrt 7 }}\theta$
Substituting the value $\theta$ from the equation (7), $\theta = \arctan \dfrac{2}{{\sqrt 7 }}\left( {x - \dfrac{1}{2}} \right)$ we get,
$\Rightarrow \dfrac{2}{{\sqrt 7 }}\arctan \left( {\dfrac{2}{{\sqrt 7 }}\left( {x - \dfrac{1}{2}} \right)} \right)$
Hence The equation (4) becomes,
$\Rightarrow \left( {x - \dfrac{1}{2}} \right)\ln ({x^2} - x + 2) - 2x + \dfrac{7}{2} \cdot \dfrac{2}{{\sqrt 7 }}\arctan \left( {\dfrac{2}{{\sqrt 7 }}\left( {x - \dfrac{1}{2}} \right)} \right)$
$\Rightarrow \left( {x - \dfrac{1}{2}} \right)\ln ({x^2} - x + 2) - 2x + \sqrt 7 \arctan \left( {\dfrac{{2x - 1}}{{\sqrt 7 }}} \right)$

Hence $\int {\ln ({x^2}} - x + 2)dx = \left( {x - \dfrac{1}{2}} \right)\ln ({x^2} - x + 2) - 2x + \sqrt 7 \arctan \left( {\dfrac{{2x - 1}}{{\sqrt 7 }}} \right)$.

Note:
Remember to make use of method of integration of parts which is given by,
$\int {udv = uv - \int {vdu} }$.
It is important to make the right choice of u and v. Students may go wrong in this. We have to make a clever choice so that the equation gets simplified.
Also sometimes the terms will be complicated while integrating. In this situation we make use of a substitution method or any other which gives us the required solution.