Question

Evaluate the Equation\[\int {\frac{{\sqrt {1 + {x^2}} }}{x}} \,dx\]

Short answer

The value of the given integral is

\[\left[ {\frac{1}{2}ln\left( {\frac{1}{{\sqrt {{x^2} + 1} }} - 1} \right) + \sqrt {{x^2} + 1} - \frac{1}{2}ln\left( {\frac{1}{{\sqrt {{x^2} + 1} }} + 1} \right)} \right] + c\]

Suppose\[x = tan\,t\]and convert the given integral in terms of\[t\]and then integrate it with respect to\[t\]. After integration, again integration, again put the value of \[t\] as \[t = ta{n^{ - 1}}x\].

Step-by-step solution

Put \[x = tan\,t\] in the given integral.

Let \[x = tan\,t\]

\[dx = \frac{1}{{co{s^2}t}}dt\]

The integral becomes:

Put \[cos\,t = u\] in the given integral.

Let \[cos\,t = u\]

\[ - sin\,t\,dt = du\]

Now, the integral becomes,

\[ = \int {\frac{{ - du}}{{\left( {1 - {u^2}} \right){u^2}}}} \]

\[ = - \int {\frac{{du}}{{\left( {1 - u} \right)\left( {1 + u} \right){u^2}}}} \]

Apply partial faction.

On applying partial fraction, the integral would become,

\[\begin{aligned}{l} = - \int {\left\{ { - \frac{1}{{2\left( {u - 1} \right)}} + \frac{1}{{{u^2}}} + \frac{1}{{2\left( {u + 1} \right)}}} \right\}du} \\ = \left[ {\frac{1}{2}ln\left( {u - 1} \right) + \frac{1}{u} - \frac{1}{2}ln\left( {u + 1} \right)} \right] + c\end{aligned}\]

Put \[u = cost\] in the values of the integral.

\[ = \left[ {\frac{1}{2}ln\left( {cos\,t - 1} \right) + \frac{1}{{cos\,t}} - \frac{1}{2}ln\left( {cos\,t + 1} \right)} \right] + c\]

Convert the value of the integral in \[x\] using \[x = tan\,t\].

As \[tan\,t = \frac{x}{1}\]

\[\therefore cos\,t = \frac{1}{{\sqrt {{x^2} + 1} }}\]

Putting \[cos\,t = \frac{1}{{\sqrt {{x^2} + 1} }}\] in the value of the integral, we get

\[ = \left[ {\frac{1}{2}\ln \left( {\frac{1}{{\sqrt {{x^2} + 1} }} - 1} \right) + \sqrt {{x^2} + 1} - \frac{1}{2}\ln \left( {\frac{1}{{\sqrt {{x^2} + 1} }} + 1} \right)} \right] + c\]

Hence, \[\int {\frac{{\sqrt {1 + {x^2}} }}{x}dx} = \left[ {\frac{1}{2}ln\left( {\frac{1}{{\sqrt {{x^2} + 1} }} - 1} \right) + \sqrt {{x^2} + 1} - \frac{1}{2}ln\left( {\frac{1}{{\sqrt {{x^2} + 1} }} + 1} \right)} \right] + c\]