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