Question

Make a substitution to express the integrand of u rational function and then evaluate the integral.

\(\int\limits_0^1 {\frac{1}{{1 + \sqrt(3){x}}}} dx\) (let \(u = \sqrt(3){x}\) )

Short answer

The value of the given integral is: \(\frac{{ - 3}}{2} + 3\ln 2\)

Step-by-step solution

Step-1: Rewrite the integral ad obtain the partial fraction

Let, \(u = \sqrt(3){x}\) \(du3{u^3} = dx\)

\({u^3} = x\) \(3{u^2}du = dx\)

The limits are \(x \to 0\)to 1 implies, \(u \to 0\)to 1

Rewrite the integral as \(\int\limits_0^1 {\frac{1}{{1 + \sqrt(3){x}}}} dx = 3\int\limits_0^1 {\frac{{{u^2}}}{{1 + u}}du} \)

Apply the long division to obtain the partial fraction as shown below:-

\(\begin{array}{l}u + 1\sqrt {{u^2}} \\\end{array}\)

This implies, \({u^2} = u\left( {u + 1} \right) - u\)

Substitute in the integral

\(\int\limits_0^1 {\frac{1}{{1 + \sqrt(3){x}}}dx = 3\int\limits_0^1 {\frac{{u\left( {u + 1} \right) - u}}{{u + 1}}} } du\)

=\(3\int\limits_0^1 {(\frac{{u\left( {u + 1} \right)}}{{\left( {u + 1} \right)}}} - \frac{u}{{u + 1}})du\)

=\(3\int\limits_0^1 {\left( {u - \frac{{u + 1 - 1}}{{u + 1}}} \right)} du\)

Step-2:- Find the integral

\(\int\limits_0^1 {\frac{1}{{1 + \sqrt(3){x}}}} dx = 3\int\limits_0^1 \begin{array}{l}\left( {u - \frac{{u + 1}}{{u + 1}} + \frac{1}{{u + 1}}} \right)du\\\end{array} \)

=\(3\int\limits_0^1 {\left( {u - 1 + \frac{1}{{u + 1}}} \right)} du\)

= \(3(\frac{{{u^2}}}{2} - u + \ln \left( {u + 1} \right))_0^1 = 3(\frac{{{{\left( 1 \right)}^2}}}{2} - \left( 1 \right) + \ln |t + 1|) - 3(\frac{{{{\cos }^2}}}{2} - \left( 0 \right) + \ln \left( {0 + 1} \right))\)

= \(\frac{3}{2} - 3 + 3\ln 2 = \frac{{ - 3}}{2} + \ln 2\)

Hence, the solution of the given integral is given as follows:

\(\)\(\int\limits_0^1 {\frac{1}{{1 + \sqrt(3){x}}}} dx = \frac{{ - 3}}{2} + \ln 2\)