Question

How do you integrate $\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}dx$ ?

Answer 1

Here, we are asked to integrate $\int {\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}} dx$. For this, we will first consider the denominator and complete the square. After that, we will substitute an appropriate value and convert the term as the function of another variable to integrate the given term easily.

Complete step by step answer:
We have $\int {\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}} dx$. First, we will complete the square at the denominator. Our denominator is ${x^2} - 4x + 3$. We will add and subtract 4 to this polynomial to make it complete square.
${x^2} - 4x + 3 = {x^2} - 4x + 4 - 4 + 3 = {x^2} - 4x + 4 - 1$
We know that ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$
Therefore, here we can write ${x^2} - 4x + 4 = {\left( {x - 2} \right)^2}$
Thus the polynomial now becomes ${\left( {x - 2} \right)^2} - 1$

\Rightarrow\int {\dfrac{1}{{\sqrt {{{\left( {x - 2} \right)}^2} - 1} }}} dx \\\ $$ Now, we will substitute $x - 2 = \sec t$ differentiating both the sides, we get $dx = \sec t\tan tdt$ Now, we will put these values in the main term. $$\int {\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}} dx \\\ \Rightarrow \int {\dfrac{1}{{\sqrt {{{\left( {x - 2} \right)}^2} - 1} }}} dx \\\ \Rightarrow \int {\dfrac{{\sec t\tan t}}{{\sqrt {{{\sec }^2}t - 1} }}} dt \\\ $$ We know that ${\sec ^2}t - 1 = {\tan ^2}t$ $$\Rightarrow \int {\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}} dx \\\ \Rightarrow \int {\dfrac{1}{{\sqrt {{{\left( {x - 2} \right)}^2} - 1} }}} dx \\\ \Rightarrow \int {\dfrac{{\sec t\tan t}}{{\sqrt {{{\sec }^2}t - 1} }}} dt \\\ \Rightarrow\int {\dfrac{{\sec t\tan t}}{{\sqrt {{{\tan }^2}t} }}} dt \\\ \Rightarrow \int {\dfrac{{\sec t\tan t}}{{\tan t}}} dt \\\ \Rightarrow\int {\sec t} dt \\\ $$ Also, we have taken $\sec t = x - 2$ We know that ${\sec ^2}t - 1 = {\tan ^2}t \\\ \Rightarrow {\tan ^2}t = {\left( {x - 2} \right)^2} - 1 \\\ \Rightarrow \tan t = \sqrt {{x^2} - 4x + 3} \\\ $ Putting both these values, we will get our final answer. $$\int {\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}} dx \\\ \Rightarrow \int {\dfrac{1}{{\sqrt {{{\left( {x - 2} \right)}^2} - 1} }}} dx \\\ \Rightarrow \int {\dfrac{{\sec t\tan t}}{{\sqrt {{{\sec }^2}t - 1} }}} dt \\\ \Rightarrow \int {\dfrac{{\sec t\tan t}}{{\sqrt {{{\tan }^2}t} }}} dt \\\ \Rightarrow \int {\dfrac{{\sec t\tan t}}{{\tan t}}} dt \\\ \Rightarrow \int {\sec t} dt \\\ \Rightarrow\ln \left| {\sec t + \tan t} \right| + c \\\ \therefore \ln \left| {x - 2 + \sqrt {{x^2} - 4x + 3} } \right| + c \\\ $$ **Hence, our final answer is: $$\int {\dfrac{1}{{\sqrt {{x^2} - 4x + 3} }}} dx = \ln \left| {x - 2 + \sqrt {{x^2} - 4x + 3} } \right| + c$$.** **Note:** Here, we have considered $\sqrt {{{\tan }^2}t} $ is equal to $\tan t$ which is a positive value and have not considered $ - \tan t$ which is negative value. There is a reason behind this. We know that the given integral can only be defined when the term given in the square root is greater than zero which means ${x^2} - 4x + 3$ must be greater than zero. We have substituted $\sec t$ in place of $x - 2$ after completing the square. Therefore, the value of $t$ will be in the range from $0$ to $\dfrac{\pi }{2}$.We have seen that the value of the polynomial ${x^2} - 4x + 3$ will be $\sqrt {{{\tan }^2}t} $. Now, we know that in the range from $0$ to $\dfrac{\pi }{2}$, which is the first quadrant, the value of tangent function is always positive. That is why we have taken $\sqrt {{{\tan }^2}t} $ is equal to $\tan t$.