The integral of one divided by the square root of $x$ square plus $a$ square with respect to $x$ is evaluated by the following integration rule of reciprocal function in irrational form.
$\displaystyle \int{\dfrac{1}{\sqrt{x^2+a^2}}}\,dx$ $\,=\,$ $\log_{e}{\big|x+\sqrt{x^2+a^2}\big|}+c$
It is time to learn how to prove the integral of one divided by square root of $x$ square plus $a$ square is equal to the natural logarithm of $x$ plus square root of $x$ square plus $a$ square, plus the constant of integration $c$.
The one divided by square root of $x$ square plus $a$ square is an irrational function in rational form. It is tough to find its integration directly but the integral of this function can be calculated by changing the variable $x$.
Let’s consider a right triangle, assume that the angle of a right angled triangle is theta, the length of opposite side is denoted by $x$ and the length of adjacent side is denoted by a literal $a$. Now, the tan of angle theta can be written in terms of a ratio in mathematics.
$\tan{\theta}$ $\,=\,$ $\dfrac{x}{a}$
$\implies$ $a \times \tan{\theta}$ $\,=\,$ $x$
$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $a \times \tan{\theta}$
The above equation expresses that the value of $x$ can be obtained by the product of a constant $a$ and a trigonometric function tan of angle theta. In other words, the value of $x$ is defined in terms of an angle theta. So, differentiate the expressions on both sides of the equation with respect to theta.
$\implies$ $\dfrac{d}{d\theta}\,x$ $\,=\,$ $\dfrac{d}{d\theta}\,(a \times \tan{\theta})$
On left hand side of the equation, the derivative of $x$ with respect to theta cannot be evaluated. So, it is written as derivative of $x$ with respect to theta. On right hand side of the equation, $a$ is a constant in the product. So, it can be released from the differentiation as per the constant multiple rule of the derivatives.
$\implies$ $\dfrac{dx}{d\theta}$ $\,=\,$ $a \times \dfrac{d}{d\theta}\,\tan{\theta}$
The derivative of tan of angle theta with respect to theta is the square of secant of angle theta as per the derivative rule of tan function.
$\implies$ $\dfrac{dx}{d\theta}$ $\,=\,$ $a \times \sec^2{\theta}$
$\implies$ $\dfrac{dx}{d\theta}$ $\,=\,$ $a\sec^2{\theta}$
$\implies$ $dx$ $\,=\,$ $a\sec^2{\theta} \times d\theta$
$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $a\sec^2{\theta}\,d\theta$
The value of the differential $dx$ can be now evaluated by the product of $a$, secant square of angle theta and the differential element $d\theta$. Now, change the entire given irrational function in the integration by substituting the value of $x$ and the differential element $dx$ with their corresponding values.
$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{x^2+a^2}}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{\sqrt{(a\tan{\theta})^2+a^2}}}\,a\sec^2{\theta}\,d\theta$
The integral of a function in terms of $x$ is converted as the integral of a function in terms of theta by the change of variable technique. Now, let us focus on simplifying the trigonometric function in irrational form.
$=\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{a^2\tan^2{\theta}+a^2}}}\, \times a\sec^2{\theta} \times d\theta$
Two functions are multiplied inside the integral operation. Now, multiply both of them to find their product by the multiplication of the fractions.
$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times a\sec^2{\theta}}{\sqrt{a^2\tan^2{\theta}+a^2}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{a^2\tan^2{\theta}+a^2}}}\,d\theta$
It is time to simplify the expression in the denominator of the rational function. The square of constant $a$ is a common factor in the two terms of expression under the square root. So, it can be taken out common from them.
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{a^2 \times \tan^2{\theta}+a^2 \times 1}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{a^2 \times 1+a^2 \times \tan^2{\theta}}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{a^2 \times (1+\tan^2{\theta})}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{a^2 \times (\sec^2{\theta})}}}\,d\theta$
According to the Pythagorean identity of secant and tangent functions, the sum of one plus square of tan function is equal to the square of secant function.
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{a^2 \times \sec^2{\theta}}}}\,d\theta$
Now, let us focus on simplifying the irrational function further to simplify the function to the maximum level.
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{\sqrt{(a\sec{\theta})^2}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec^2{\theta}}{a\sec{\theta}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\dfrac{\cancel{a\sec^2{\theta}}}{\cancel{a\sec{\theta}}}}\,d\theta$
$=\,\,\,$ $\displaystyle \int{\sec{\theta}}\,d\theta$
According to the proof of the integral of secant function, the integral of the secant of angle theta with respect to theta can be derived as the natural logarithm of the secant of angle theta and tan of angle theta, plus the integral constant $c_1$.
$\implies$ $\displaystyle \int{\sec{\theta}}\,d\theta$ $\,=\,$ $\log_{e}{\big|\sec{\theta}+\tan{\theta}\big|}$ $+$ $c_1$
The integral of secant of angle theta is successfully evaluated with respect to theta but the given function is defined in terms of $x$. So, let us try to find the integral in terms of $x$.
$=\,\,\,$ $\log_{e}{\big|\sec{\theta}+\tan{\theta}\big|}$ $+$ $c_1$
According to the Pythagorean identity of secant and tangent functions, the secant of angle theta can be written in terms of tan of angle theta.
$\sec{\theta}$ $\,=\,$ $\sqrt{1+\tan^2{\theta}}$
Now, replace the value of secant function in the natural logarithmic expression.
$=\,\,\,$ $\log_{e}{\big|\sqrt{1+\tan^2{\theta}}+\tan{\theta}\big|}$ $+$ $c_1$
The natural logarithmic expression is purely written in terms of tan of angle theta. It is time to eliminate the tan of angle theta by its actual value in $x$.
$=\,\,\,$ $\log_{e}{\Bigg|\sqrt{1+\Big(\dfrac{x}{a}\Big)^2}+\dfrac{x}{a}\Bigg|}$ $+$ $c_1$
The square is a common exponent for both $x$ in numerator and a in denominator. So, it can be distributed to both of them by the power of a quotient rule.
$=\,\,\,$ $\log_{e}{\Bigg|\sqrt{1+\dfrac{x^2}{a^2}}+\dfrac{x}{a}\Bigg|}$ $+$ $c_1$
Add the quotient of $x$ square plus $a$ square to the number one by the addition of the fractions, inside the logarithm under the square root to simplify the mathematical expression in logarithmic form.
$=\,\,\,$ $\log_{e}{\Bigg|\sqrt{\dfrac{a^2+x^2}{a^2}}+\dfrac{x}{a}\Bigg|}$ $+$ $c_1$
$=\,\,\,$ $\log_{e}{\Bigg|\sqrt{\dfrac{x^2+a^2}{a^2}}+\dfrac{x}{a}\Bigg|}$ $+$ $c_1$
According to the power rule of a quotient, the square root can be distributed to both $x$ square plus $a$ square and $a$ square.
$=\,\,\,$ $\log_{e}{\Bigg|\dfrac{\sqrt{x^2+a^2}}{\sqrt{a^2}}+\dfrac{x}{a}\Bigg|}$ $+$ $c_1$
$=\,\,\,$ $\log_{e}{\Bigg|\dfrac{\sqrt{x^2+a^2}}{a}+\dfrac{x}{a}\Bigg|}$ $+$ $c_1$
The fractions inside the logarithmic function can be added by the adding rule of fractions.
$=\,\,\,$ $\log_{e}{\Bigg|\dfrac{\sqrt{x^2+a^2}+x}{a}\Bigg|}$ $+$ $c_1$
$=\,\,\,$ $\log_{e}{\Bigg|\dfrac{x+\sqrt{x^2+a^2}}{a}\Bigg|}$ $+$ $c_1$
The logarithm of a quotient can be written as difference of two logarithmic functions by the quotient rule of logarithms.
$=\,\,\,$ $\log_{e}{\big|x+\sqrt{x^2+a^2}\big|}$ $-$ $\log_{e}{|a|}$ $+$ $c_1$
The value natural logarithm of a constant $a$ is also a constant and $c_1$ already represents a constant. So, the negative natural logarithm of a constant $a$ plus the constant $c_1$ can be denoted by a constant $c$ simply, and it is called the constant of integration.
$=\,\,\,$ $\log_{e}{\big|x+\sqrt{x^2+a^2}\big|}$ $+$ $c$
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